# Calculate the pH of a buffer solution that is simultaneously 0

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```					                                             BUFFERS

Definition – solutions which are resistant to changes in their pH when moderate amounts of acid
or base are added to them.

An example of a non buffered solution is pure H2O.

Example: adding 10 mL of 0.100 M HCl to 1.00 L of pure water changes the pH from 7 to about
3.

Typical Buffer System – a mixture of a weak acid and its salt or a weak base and its salt
(depending on whether we want the resultant buffer pH to be above or below 7).

Why A System Like This Is Protected Against Changes In pH?

Consider a buffer consisting of HC2H3O2 and NaC2H3O2, acetic acid and sodium acetate.

a) If H+ is added, the acetate ion, C2H3O2-, "gobbles it up."

b) If OH- is added, the HC2H3O2 "gobbles it up."

HC2H3O2  H+ + C2H3O2-

For any weak acid HB  H+ + B-

Ka = [H+] [B-] Solving for [H+]
[HB]

[H+] = Ka  [HB] = Ka  # mol HB
[B-]         #mol B-

You should solve most of these problems as a typical equilibrium-Ka problem. I can never
remember these above or what we will later call the Henderson-Hasselblach Equation. You
should probably solve these by using the method as the next problem shows.
Practice Problem:

Calculate the pH of a buffer solution that is simultaneously 0.100 M HC2H3O2 and 0.100 M
NaC2H3O2. [You need to look up the Ka for acetic acid.] A buffer like this is referred to as a
"symmetrical" buffer because the concentrations of the two ingredients are the same. Solve it as a
regular Ka problem.

HC2H3O2  H+ + C2H3O2-

I                          0.100 M                     0                          0.100M

C                          -x                          x                          x

E                          -x                          x                          0.100 + x

Ka = [H+][C2H3O2-] = (x)(0.100 + x)         Assume x is small
[HC2H3O2]        [0.100-x]

1.8 x 10-5 = (x)(0.100)                   x = 1.8 x 10-5
(0.100)

pH = - log[1.8 x 10-5] = 4.74

Three Common Methods Of Preparing A Buffer Solution:

1. Mix a weak acid (or base) and its salt. In choosing an appropriate acid or base system,
you consider what the desired pH of the buffer is supposed to be. A symmetrical buffer
will buffer at a pH = pKa of the weak acid used (or pOH = pKb of the weak base used).
THEREFORE, you select an acid or base that has a pKa (or a pKb) close to the pH (or
pOH) you want. You can vary the relative amounts of the two components of the buffer
to get the actual pH you want.
2. Partially neutralize a solution of a weak acid by using a strong base. This works because
the acid molecules that are neutralized are converted to ions, e.g.,

HC2H3O2 + OH-  C2H3O2- + H2O

If only a part of the acid molecules is neutralized, we create a solution containing a weak
acid and its salt.

You can accomplish #2 by ½ Neutralizing a weak acid.

HB  H+ + B-

[H+] = Ka x [HB] / [B-]

If you ½ neutralize the Acid [HB]= [B-] & [H+] = Ka
3. Add a strong acid to a solution of the salt of a weak acid using fewer moles of the strong
acid than you have of the salt of the weak acid.

The Reation is: H+ + C2H3O2-  HC2H3O2

Practice Problem:

250.0 mL of 0.500 M NaOH are added to 350.0 mL of 0.900 M HC2H3O2. Calculate the pH of
the resultant buffer.

HC2H3O2  H+ + C2H3O2-                Ka= 1.8x10-5

I We want to get these – we want INT. # moles of HC2H3O2 added.

C

E

(.3500L) (.900 Mole / L) = .0315 Mole HC2H3O2 # moles of OH- added.

(.3500L) (.500 Mole / L) = .0125 Mole of OH-

SO

OH- + HC2H3O2  C2H3O2- + H2O

# Moles C2H3O2-        Formed = .0215 Moles C2H3O2-

# Moles HC2H3O2         Remaining .0315 - .0215 = .190 Moles

Total Volume = 250.0 + 350.0 = 600.0 ML

HC2H3O2                      H+        +            C2H3O2-

INT          I              0.190 / .6L                    0                      0.125 / .6L

∆           C               -x                             x                      x

EQU         E               0.190 / .6 – x                 x                      0.125 / .6 + x

Ka = 1.8 x 10-5 = (x) (.125/.6 + x) / (.190 /.6 –x) Assume x is small.

X = 2.7 x 10-5        So pH = 4.57

EFFECT OF ADDING A STRONG BASE OR ACID TO A BUFFER:

HB + OH-  H2O + B-
[ALL SPECIES ARE (aq).]

B- + H+  HB

As a result of these reactions, H+ and OH- ions are consumed, so the pH does not change
dramatically.

PRACTICE PROBLEM: To our original buffer in problem #1, 100.0 mL of 0.100 M HCl
are added to 1.00 L of a buffer solution that is 0.100 M in both acetic acid and sodium
acetate. Calculate the pH of the resulting solution.

Moles of HC2H3O2 and C2H3O2- before

(1.00L) (.100 Mol/L) = .100 Moles each

(.100 L) (.100 Mol/L) =.0100 moles of H+

Reaction that takes place

H+ + C2H3O2-  HC2H3O2

After this reaction

# Moles C2H3O2- = .100 - .0100 = .090 Mole

# Moles HC2H3O2 = .100 + .0100 = .0110 Mole

HC2H3O2  H+ + C2H3O2-

Total Vol = 1.10 L

INT    I                0.110/1.10 Mole                        0    0.090 Mol /1.10 L

∆      C                -x                                     x    x

EQU E                   0.110 Mol / 1.10 – x                   x    0.090 Mol / 1.10L + x

ka = [H+][C2H3O2-] = (x)(0.090/1.10L + x)
[HC2H3O2]       (0.110/1.10L – x)

Drop x because it is small
1.8 x 10-5 = (x)(0.090/1.10) = 2.2 x 10-5 M
(0.110/1.10)

pH = + 4.66

Therefore not much of a change

HENDERSON-HASSELBACH EQUATION

There is an equation that many teachers like to utilize when working with buffer solution
problems called the Henderson-Hasselbach equation [see p. 512], derived from the
equilibrium constant expression, Ka or Kb.

pH = pKa + log[B-]
[HB]

I fail to understand why anyone really wants to use this equation. It just seems like one
more thing for students to memorize and it would seem relatively easy for students to
mix up the equation and get it "upside down." I have always wondered how Henderson
and Hasselbalch (I assume they are two different individuals) managed to get their
names in our textbooks for something that looks so trivial. Anyway, here is how the
equation is derived. On the second thought forget it.

MODEL-PRACTICE PROBLEM:

Prepare 500. mL (3 sig figs!) of a buffer solution that has a pH of 7.00 by using only
0.500 M hypochlorous acid, HClO, and 0.350 M sodium hypochlorite, NaClO. The Ka for
HClO = 2.95 x 10-8.

HClO  H+ + ClO-               Ka = 2.95 x 10-8

Since the pH is 7.00, the [H+] = 10-7. We want to know the volumes of HClO and NaClO
needed.

Ka= [H+] [ClO-] = 2.95 x 10-8 = (10-7) [ClO-]
[HClO]                      [HClO]

Ka    = [ClO-]      = 0.295
[H+]     [HClO]

Let x = volume in liters of 0.500 M HClO required. Then (.500-x) = volume in liters of
0.350 M NaClO required.

[ClO-] = (.500-x) (.350)/0.500 L = .295
[HClO]      (x) (.500)/0.500L

0.175 - 0.350 x = 0.295
.500 x
Solving for x, x = 0.352 L so we will use 352 mL of 0.500 M HClO and (500 - 352) = 148
mL of 0.350 M NaClO.

BUFFER CAPACITY

This refers to the amount of acid or base that can be added to a buffer solution before its
pH changes "appreciably." It can be demonstrated that the most effective buffer exists
when the concentrations of both the weak acid and its salt (or weak base and its salt)
are high and are equal to each other, a symmetrical buffer. A buffer, therefore, has its
maximum capacity when pH = pKa (or pOH = pKb). Whenever the ratio of salt to weak
acid or base becomes either less than 0.10 or more than 10.0, the buffer begins to lose
its effectiveness. Thus, a buffer is most effective in a range about one pH unit on either
side of the pKa (or pKb) of the weak acid or base present in the buffer system.

HUMAN BLOOD AS A BUFFER

It is absolutely essential that the pH of human blood and intercellular fluids remain fairly
constant, because the functioning of many enzymes that operate in human biochemistry
is often sharply pH dependent.

The pH of human blood is 7.4. Severe illness or death results if the pH varies even a few
pH units from this value for any sustained period of time.

Acidosis (low blood pH) can result from heart failure, kidney failure, diabetes, persistent
diarrhea, or a long-term high protein diet. Prolonged, intense exercise can cause a
temporary condition of acidosis.

Alkalosis (high blood pH) can be caused by severe vomiting, hyperventilation, or
exposure to high altitudes.

Studies performed on climbers who reached the summit of Mount Everest (29,028 ft)
without supplemental oxygen had arterial blood with a pH of 7.7-7.8. Extreme
hyperventilation required to compensate for the low partial pressure of O2 (about 43 mm
Hg) caused this condition.

Blood is a very complex buffer. The major buffering system involves the carbonic acid-
bicarbonate ion system (actually CO2 and HCO3-). Interestingly enough, the ratio of
HCO3- to H2CO3 is about 10:1, which would seem to place this buffer outside the range
of its maximum buffering capacity. The situation is rather complex, but this seems useful
because:

a) the need to neutralize excess acid (lactic acid produced by exercise) is generally
greater than the need to neutralize excess base. The high proportion of HCO3- helps in
this regard.

b) If additional H2CO3 is needed to neutralize excess alkalinity, CO2(g) in the lungs can be
reabsorbed to increase the H2CO3 content of the blood.
c) Other components, such as H2PO4-/HPO42- and some plasma proteins contribute to
maintaining the pH at 7.4.

THEORY OF OPERATION OF INDICATORS

Indicators are not "magical" substances. They are simply weak acids (usually) which
happen to have one additional and useful property--the color of the molecular (acid) form
of the indicator, HIn, is a different color from the ionic (basic) form of the indicator, In-.

Since they are weak acids:

HIn  H+ + In- and Ka = [H+] [In-]
[HIn]

Rearranging, [HIn] = [H+]
[In-]    Ka

This shows that the 2 forms of the indicator depend upon the [H+] in the solution. Since
the [H+] changes by several powers of 10 near the equivalence point of a titration, the
ratio of the two forms of the indicator changes by this same factor. That is why the color
changes so rapidly and dramatically.

WHY AN INDICATOR CHANGES COLOR WHEN THE pH OF THE SOLUTION
EQUALS THE pKa OF THE INDICATOR:

When an indicator is just ready to change color, [HIn] = [In-].

This means that

[H+] [In-] = Ka   so [H+] = Ka       and pH = pKa
[HIn]

TITRATIONS: Titration of a strong (monoprotic) acid by a strong base

Ex. Consider the titration of 25.00 mL of 0.1000 M HCl by 0.1000 M NaOH. Calculate the
pH after 10.00 mL of base has been added.

Initial # Mol H+
(0.02500L)(0.1000 Mol/L) = 0.002500 Mol H+

(0.01000L)(0.1000 Mol/L) = 0.001000 Mol OH-

After Reaction

#Mol H+ remaining = 0.002500-0.00100
= 0.001500 Mol H+
[H+] = 0.001500 Mol = 0.0426M
0.03500L

pH = - log(0.0426) = 1.3680

DATA for pH of the solution vs volume of NaOH added:

Titration curve for a strong acid by a strong base--25.00 mL of 0.1000 M HCl by 0.1000
M NaOH.

mL of NaOH                pH
0.00                  1.00
10.00                  1.37
20.00                  1.95
22.00                  2.19
24.00                  2.69
25.00                  7.00
26.00                  11.29
28.00                  11.75
30.00                  11.96
40.00                  12.36
50.00                  12.52

TITRATION OF A WEAK MONOPROTIC ACID WITH A STRONG BASE

Consider the titration of 25.00 mL of 0.1000 M HC2H3O2 with 0.1000 M NaOH. Let’s look
at sample calculations of pH of the solution at various points during the titration:

a. For the original solution find pH before any base added. What is the chemistry
you need to solve this?

HC2H3O2                 H+                      A-

I                       0.1000                   0                       0

C                       -x                       x                       X

E                       0.1000 – x               x                       x
Ka = 1.8 x 10-5 = x2                      small x
0.1000 – x

x = 1.3 x 10-3           So      pH = 2.87

b) At the equivalence point of the titration: at this point, the neutralization reaction has
"gone to completion," and what we have is essentially a solution of sodium acetate,
NaC2H3O2. Since the acetate ion is the conjugate base of a weak acid, it will hydrolyze
according to the following reaction:

C2H3O2- + H2O  HC2H3O2 + OH-

Ka of acetic acid = 1.8 x 10-5

AT THE END POINT

Moles C2H3O2- = 0.02500 L x 0.1000 Mole/L
= 0.002500 Moles

Vol Total = 50.00 mL

[C2H3O2-]int = 0.002500 Moles = 0.05000 M
0.0500 L

C2H3O2- + H2O                    HC2H2O2 +                          OH-

I                         0.05000M                        0                                    0

C                         -x                              x                                    x

E                         0.05000 – x                     x                                    x

kb = kw = 1.0 x 10-14 = 5.6 x 10-10 = [HA][OH-]
ka 1.8 x 10-5                      [A-]

5.6 x 10-10 =   x2               x = 5.3 x 10-6 M = [OH-]
0.0500 – x    small x

pOH = - log[5.3x10-6M] = 5.28

pH = 8.72

c) After 26.00 mL of base have been added find the pH:

Even though we are just past the equivalence point, we can make the assumption that
virtually all of the OH- in the solution will come from the excess, unneutralized base. The
amount coming form the hydrolysis from the acetate ion can be neglected.
Since the acid and base were of the same concentration, the original 25.00 mL of acid
required 25.00 mL of base to be neutralised. Therefore, there are 1.00 mL of base that
are in excess, and [note total volume is now 51 mL]

#Moles OH- left over = (0.00100L)(0.1000 Mole/L)

= 1.00 x 10-4 Mole OH-

[OH-] = 1.00 x 10-4 Mole OH- = 1.96 x 10-3M
0.05100L

pOH = - log 1.96 x 10-3 = 2.708

pH = 14 – 2.71 = 11.292

KEY POINTS TO NOTE:

1. The initial pH is higher than in the titration of a strong acid because the weak acid is
only partially ionized.

2. There is a rather sharp increase in pH at the start of the titration.

3. Over a long section of the curve preceding the equivalence point the pH changes
quite gradually. Solutions for this portion of the curve are buffer solutions.

4. At the point of half-neutralization, pH = pKa. At the half-neutralization point, [HA] = [A-].

5. The pH at the equivalence point is greater than 7. The anion of the salt hydrolyses
because it is the conjugate base of a weak acid.

6. Beyond the equivalence point the titration curve is identical to that of a strong acid by
a strong base. In this portion of the titration the pH is established by the concentration of
unreacted OH-.

7. The steep portion of the titration curve at the equivalence point is over a relatively
short pH range (for example, from about pH 7 to 10).

8. The selection of indicators available for the titration is more limited than in a strong
acid-strong base titration.

Therefore, for a weak acid-strong base titration, you need an indicator that changes
color above pH 7. This means the pKa for the indicator is >7. The opposite for a strong
acid & weak base.

TITRATION OF A WEAK BASE WITH A STRONG ACID

Consider 50.00 mL of 1.000 M NH3 titrated with 1.000 M HCl looking at 3 stages in the
titration.
a) before any hydrochloric acid is added:

b) after 25.00 mL of the HCl has been added:

c) after 50.00 mL of acid has been added, that is, at the equivalence point:

d) after 51.00 mL of acid has been added:

TITRATION OF A POLYPROTIC WEAK ACID WITH A STRONG BASE

Consider the titration of 10.00 mL of 0.1000 M H3PO4 with 0.1000 M NaOH.

Essentially three separate neutralizations take place in successive order. The equations
for these neutralizations are respectively

H3PO4 + OH-  H2O + H2PO4-

H2PO4- + OH-  H2O + HPO42-

HPO42-+ OH-  H2O + PO43-

Corresponding to these three distinctive stages, we would expect there to be three
equivalence points. For every mole of H3PO4, 1 mole of NaOH is required to reach the
first equivalence point. At this first equivalence point, the solution is essentially just a
solution of NaH2PO4. This solution should be somewhat acidic, because the acid
ionization constant of H2PO4- (Ka = 6.34 x 10-8) is greater than the base ionization
(hydrolysis) constant, (Kb = 1.41 x 10-12).

H2PO4- + H2O  H3O+ + HPO42-            Ka = 6.34 x 10-8

H2PO4- + H2O  H3PO4 + OH-             Kb = 1.41 x 10-12

To reach the second equivalence point, 1 more mole of NaOH will be required. Now we
essentially have a solution of Na2HPO4. This solution will be somewhat basic, because
the base ionization (hydrolysis) constant of HPO42- is greater than the acid ionization
constant of HPO42-.

HPO42- + H2O  H3O+ + PO43-            Ka = 4.22 x 10-13

HPO42- + H2O  OH- + H2PO4-            Kb = 1.58 x 10-7

Although you anticipate a third equivalence point, none is present. This is because at the
third equivalence point you will have what essentially is a solution of Na3PO4. Phosphate
ion is such a strong Bronsted-Lowry base that it hydrolyses extensively. In fact the pH
may, depending on the concentration of the solution, be higher than 13, which is the pH
of the 0.1000 M NaOH that we are using in the titration. Thus, there is no way that we
can show a rapid increase in pH as add the NaOH, so no third equivalence point can be
detected.

TITRATION CURVE FOR 0.1000 M H3PO4 BEING TITRATED WITH 0.1000 M NaOH

SOLUBILITY EQUILIBRIUM: Ksp

A. Expression for Ksp: AgCl(s)  Ag+(aq) + Cl-(aq)

Ksp = 1.6 x 10-10 = [Ag+] [Cl-]

PbCl2(s)  Pb2+(aq) + 2 Cl-(aq)

Ksp = 1.7 x 10-5 = [Pb2+] [Cl-]2

B. Uses for Ksp

1. Calculation of concentration of one ion, knowing that of another. What is the [Pb2+] in
a solution in equilibrium with PbCl2 if [Cl-] = 0.020 M?

(ksp = 1.7 x 10-5) if [Cl-] = 0.020 M?

[Pb2+] = 1.7 x 10-5 = 0.042M
(2.0 x 10-2)2

2. Determination of whether precipitate will form. If Q (the ion product set up like the Ksp
expression) is less than Ksp, no precipitate will form (equilibrium not established). If Q >
Ksp, precipitate forms until Q becomes equal to Ksp.

Suppose enough Ag+ is added to a solution 0.001 M in CrO42- ion to make the
concentration of Ag+ equal to 0.001 M. Assuming the concentration of chromate remains
unchanged, will silver chromate precipitate? The Ksp = 2 x 10-12.

Ag2CrO4  2Ag+ + CrO42-

Q = (oring. Conc. Ag+)2 x (orig. conc. CrO42-)

= (1 x 10-3)2 x (1 x 10-3) = 1 x 10-9 > ksp so precipitate forms

Relationship between Ksp and solubility, s

Equilibrium concentration in solution
solid              Ksp expression                cation       anion   Relation between Ksp and
s
AgCl                  [Ag+] [Cl-]                  s           s             Ksp = x2 and
s = (Ksp)1/2
CoS                      [Co2+] x [Cl-]                  s               s                 Ksp = s2;

s = (Ksp)1/2
PbI2                      [Pb2+] x [I-]2                 s           2s                    Ksp = 4s2

s = (Ksp/4)1/3
Ag2CrO4                [Ag+]2 x [CrO43-]                   2s              s                 Ksp = 4s2

s = (Ksp/4)1/3
Al(OH)3                    [Al3+] x [OH-]3                 s           3s                   Ksp = 27s4

s = (Ksp/27)1/4

3. Determination of solubility

PbCl2(s) in pure water

PbCl2(s)                  Pb2+(aq) +         2Cl-(aq)
s                          s                  2s

Ksp = [Pb2+(aq)] [Cl-(aq)] 2       next plug in!

4s3 = Ksp = 1.7 x 10-5

So s = 1.6 x 10-2M

b) In solution containing a common ion: example solubility of PbCl2(s) in 0.100 M HCl.

PbCl2(s)                                  Pb2+(aq) +                              2Cl-(aq)

s                                          s                                       (2s + 0.10) about equal to
0.10

Or solving this using Ksp s(0.10)2 = 1.7 x 10-5
s= 1.7 x 10-3M

WHAT IS THE SOLUBILITY OF X3Y2 IN TERMS OF Ksp?

X3Y2  3X2+ + 2Y-3                        Ksp = [X2+]3[Y3-]2

Let 3s = [X2+]                            2s = [Y3-] so

Ksp = (3s)3(2s)2 = 27s3 x 4s2 = 108s5
Or s = (Ksp/108)1/5

Solving Acid-Base Problems:

1) List the major species in the solution.

2) Look for reactions that can be assumed to go to completion, for example, a strong
acid dissociating or H+ reacting with OH-.

3) For a Rx that can be assumed to go to completion:

a) determine the concentration of the products.

b) write down the major species in solution after the Rx.

4) Look at each major component of the solution and decide if it is an acid or a base.

5) Pick the equilibrium that will control the pH. Use known values of the dissociation
constants (Ka and Kb) for the various species to help decide on the dominant equilibrium.
Which has a larger value?

a) Write the equation for the Rx and the equilibrium expression.

b) Compare the initial concentrations (assuming the dominant equilibrium has not yet
occurred, that is, no acid dissociation, etc.)

c) Define x.

d) Compute the equilibrium concentrations in terms of x.

e) Substitute the concentrations into the equilibrium expression and solve for x.

f) Check the validity of the approximation.

g) Calculate the pH and other concentrations as required.
pH (TITRATION) CURVES

titrations.

The equivalence point of a titration

Sorting out some confusing terms

When you carry out a simple acid-base titration, you use an indicator
to tell you when you have the acid and alkali mixed in exactly the right
proportions to "neutralise" each other. When the indicator changes
color, this is often described as the end point of the titration.

In an ideal world, the color change would happen when you mix the
two solutions together in exactly equation proportions. That particular
mixture is known as the equivalence point.

For example, if you were titrating sodium hydroxide solution with
hydrochloric acid, both with a concentration of 1 mol dm-3, 25 cm3 of
sodium hydroxide solution would need exactly the same volume of the
acid - because they react 1 : 1 according to the equation.

In this particular instance, this would also be the neutral point of the
titration, because sodium chloride solution has a pH of 7.

But that isn't necessarily true of all the salts you might get formed.

For example, if you titrate ammonia solution with hydrochloric acid,
you would get ammonium chloride formed. The ammonium ion is
slightly acidic, and so pure ammonium chloride has a slightly acidic
pH.

That means that at the equivalence point (where you had mixed the
solutions in the correct proportions according to the equation), the
solution wouldn't actually be neutral. To use the term "neutral point" in

Similarly, if you titrate sodium hydroxide solution with ethanoic acid, at
the equivalence point the pure sodium ethanoate formed has a slightly
alkaline pH because the ethanoate ion is slightly basic.
To summarise:

   The term "neutral point" is best avoided.
   The term "equivalence point" means that the solutions have
been mixed in exactly the right proportions according to the
equation.
   The term "end point" is where the indicator changes color. As
you will see on the page about indicators, that isn't necessarily
exactly the same as the equivalence point.

Note: You can find out about indicators by following this
You should read the present page first though.

Simple pH curves

All the following titration curves are based on both acid and alkali
having a concentration of 1 mol dm-3. In each case, you start with 25
cm3 of one of the solutions in the flask, and the other one in a burette.

Although you normally run the acid from a burette into the alkali in a
other way around as well. Alternative versions of the curves have
been described in most cases.

Titration curves for strong acid v strong base

We'll take hydrochloric acid and sodium hydroxide as typical of a
strong acid and a strong base.

Running acid into the alkali
You can see that the pH only falls a very small amount until quite near
the equivalence point. Then there is a really steep plunge. If you
calculate the values, the pH falls all the way from 11.3 when you have

Note: If you need to know how to calculate pH changes
during a titration, you may be interested in my chemistry
calculations book.

Running alkali into the acid

This is very similar to the previous curve except, of course, that the pH
starts off low and increases as you add more sodium hydroxide
solution.
Again, the pH doesn't change very much until you get close to the
equivalence point. Then it surges upwards very steeply.

Titration curves for strong acid v weak base

This time we are going to use hydrochloric acid as the strong acid and
ammonia solution as the weak base.

Running acid into the alkali

Because you have got a weak base, the beginning of the curve is
obviously going to be different. However, once you have got an
excess of acid, the curve is essentially the same as before.

At the very beginning of the curve, the pH starts by falling quite quickly
as the acid is added, but the curve very soon gets less steep. This is
because a buffer solution is being set up - composed of the excess
ammonia and the ammonium chloride being formed.

Note: You can find out more about buffer solutions by
following this link. However, this is a very minor point in
the present context, and you would probably do better to
up.

Notice that the equivalence point is now somewhat acidic ( a bit less
than pH 5), because pure ammonium chloride isn't neutral. However,
the equivalence point still falls on the steepest bit of the curve. That
will turn out to be important in choosing a suitable indicator for the
titration.

Running alkali into the acid

At the beginning of this titration, you have an excess of hydrochloric
acid. The shape of the curve will be the same as when you had an
excess of acid at the start of a titration running sodium hydroxide
solution into the acid.

It is only after the equivalence point that things become different.

A buffer solution is formed containing excess ammonia and
ammonium chloride. This resists any large increase in pH - not that
you would expect a very large increase anyway, because ammonia is
only a weak base.
Titration curves for weak acid v strong base

We'll take ethanoic acid and sodium hydroxide as typical of a weak
acid and a strong base.

Running acid into the alkali

For the first part of the graph, you have an excess of sodium
hydroxide. The curve will be exactly the same as when you add
hydrochloric acid to sodium hydroxide. Once the acid is in excess,
there will be a difference.
Past the equivalence point you have a buffer solution containing
sodium ethanoate and ethanoic acid. This resists any large fall in pH.

Running alkali into the acid

The start of the graph shows a relatively rapid rise in pH but this slows
down as a buffer solution containing ethanoic acid and sodium
ethanoate is produced. Beyond the equivalence point (when the
sodium hydroxide is in excess) the curve is just the same as that end
of the HCl - NaOH graph.

Titration curves for weak acid v weak base
The common example of this would be ethanoic acid and ammonia.

It so happens that these two are both about equally weak - in that
case, the equivalence point is approximately pH 7.

Running acid into the alkali

This is really just a combination of graphs you have already seen. Up
to the equivalence point it is similar to the ammonia - HCl case. After
the equivalence point it is like the end of the ethanoic acid - NaOH
curve.

Notice that there isn't any steep bit on this graph. Instead, there is just
what is known as a "point of inflexion". That lack of a steep bit means
that it is difficult to do a titration of a weak acid against a weak base.

Note: Because you almost never do titrations with this
combination, there is no real point in giving the graph
where they are added the other way round. It isn't difficult
to work out what it might look like if you are interested -
take the beginning of the sodium hydroxide added to
ethanoic acid curve, and the end of the ammonia added
to hydrochloric acid one.

The reason that it is difficult to do these titrations is
discussed on the page about indicators.
A summary of the important curves

The way you normally carry out a titration involves adding the acid to
the alkali. Here are reduced versions of the graphs described above
so that you can see them all together.

More complicated titration curves

Adding hydrochloric acid to sodium carbonate solution

The overall equation for the reaction between sodium carbonate
solution and dilute hydrochloric acid is:

If you had the two solutions of the same concentration, you would
have to use twice the volume of hydrochloric acid to reach the
equivalence point - because of the 1 : 2 ratio in the equation.

Suppose you start with 25 cm3 of sodium carbonate solution, and that
both solutions have the same concentration of 1 mol dm-3. That means
that you would expect the steep drop in the titration curve to come
The actual graph looks like this:

The graph is more complicated than you might think - and curious
things happen during the titration.

You expect carbonates to produce carbon dioxide when you add acids
to them, but in the early stages of this titration, no carbon dioxide is
given off at all.

Then - as soon as you get past the half-way point in the titration - lots
of carbon dioxide is suddenly released.

The graph is showing two end points - one at a pH of 8.3 (little more
than a point of inflexion), and a second at about pH 3.7. The reaction
is obviously happening in two distinct parts.

In the first part, complete at A in the diagram, the sodium carbonate is
reacting with the acid to produce sodium hydrogencarbonate:

You can see that the reaction doesn't produce any carbon dioxide.

In the second part, the sodium hydrogencarbonate produced goes on
to react with more acid - giving off lots of CO2.

That reaction is finished at B on the graph.

It is possible to pick up both of these end points by careful choice of
indicator. That is explained on the separate page on indicators.

Adding sodium hydroxide solution to dilute ethanedioic acid

Ethanedioic acid was previously known as oxalic acid. It is a diprotic
acid, which means that it can give away 2 protons (hydrogen ions) to
a base. Something which can only give away one (like HCl) is known
as a monoprotic acid.

The reaction with sodium hydroxide takes place in two stages
because one of the hydrogens is easier to remove than the other. The
two successive reactions are:

If you run sodium hydroxide solution into ethanedioic acid solution, the
pH curve shows the end points for both of these reactions.

The curve is for the reaction between sodium hydroxide and
ethanedioic acid solutions of equal concentrations.

ACID-BASE INDICATORS

to choose the right one for a particular titration.

How simple indicators work

Indicators as weak acids
Litmus

Litmus is a weak acid. It has a seriously complicated molecule which
we will simplify to HLit. The "H" is the proton which can be given away
to something else. The "Lit" is the rest of the weak acid molecule.

There will be an equilibrium established when this acid dissolves in
water. Taking the simplified version of this equilibrium:

The unionized litmus is red, whereas the ion is blue.

Now use Le Chatelier's Principle to work out what would happen if you
added hydroxide ions or some more hydrogen ions to this equilibrium.

If the concentrations of HLit and Lit - are equal:

At some point during the movement of the position of equilibrium, the
concentrations of the two colors will become equal. The color you see
will be a mixture of the two.

The reason for the inverted commas around "neutral" is that there is
no reason why the two concentrations should become equal at pH 7.
For litmus, it so happens that the 50 / 50 color does occur at close to
pH 7 - that's why litmus is commonly used to test for acids and alkalis.
As you will see below, that isn't true for other indicators.

Methyl orange

Methyl orange is one of the indicators commonly used in titrations. In
an alkaline solution, methyl orange is yellow and the structure is:

Now, you might think that when you add an acid, the hydrogen ion
would be picked up by the negatively charged oxygen. That's the
obvious place for it to go. Not so!

In fact, the hydrogen ion attaches to one of the nitrogens in the
nitrogen-nitrogen double bond to give a structure which might be
drawn like this:
Note: You may find other structures for this with different
arrangements of the bonds (although always with the hydrogen
attached to that same nitrogen). The truth is that there is
delocalisation over the entire structure, and no simple picture will
show a real case where the color of a compound is drastically
changed by the presence or absence of a hydrogen ion.

You have the same sort of equilibrium between the two forms of
methyl orange as in the litmus case - but the colors are different.

You should be able to work out for yourself why the color changes
when you add an acid or an alkali. The explanation is identical to the
litmus case - all that differs are the colors.

In the methyl orange case, the half-way stage where the mixture of
red and yellow produces an orange color happens at pH 3.7 -

Phenolphthalein

Phenolphthalein is another commonly used indicator for titrations, and
is another weak acid.

In this case, the weak acid is colorless and its ion is bright pink.
Adding extra hydrogen ions shifts the position of equilibrium to the left,
and turns the indicator colorless. Adding hydroxide ions removes the
hydrogen ions from the equilibrium which tips to the right to replace
them - turning the indicator pink.

The half-way stage happens at pH 9.3. Since a mixture of pink and
colorless is simply a paler pink, this is difficult to detect with any
accuracy!

The pH range of indicators

The importance of pKind

Think about a general indicator, HInd - where "Ind" is all the rest of the
indicator apart from the hydrogen ion which is given away:

Because this is just like any other weak acid, you can write an
expression for Ka for it. We will call it Kind to stress that we are talking

Think of what happens half-way through the color change. At this point
the concentrations of the acid and its ion are equal. In that case, they
will cancel out of the Kind expression.

You can use this to work out what the pH is at this half-way point. If
you re-arrange the last equation so that the hydrogen ion
concentration is on the left-hand side, and then convert to pH and
pKind, you get:
That means that the end point for the indicator depends entirely on
what its pKind value is. For the indicators we've looked at above, these
are:

indicator         pKind

litmus               6.5

methyl orange        3.7

phenolphthalein      9.3

The pH range of indicators

Indicators don't change color sharply at one particular pH (given by
their pKind). Instead, they change over a narrow range of pH.

Assume the equilibrium is firmly to one side, but now you add
something to start to shift it. As the equilibrium shifts, you will start to
get more and more of the second color formed, and at some point the
eye will start to detect it.

For example, suppose you had methyl orange in an alkaline solution
so that the dominant color was yellow. Now start to add acid so that
the equilibrium begins to shift.

At some point there will be enough of the red form of the methyl
orange present that the solution will begin to take on an orange tint.
As you go on adding more acid, the red will eventually become so
dominant that you can no longer see any yellow.

There is a gradual smooth change from one color to the other, taking
place over a range of pH. As a rough "rule of thumb", the visible
change takes place about 1 pH unit either side of the pKind value.

The exact values for the three indicators we've looked at are:

indicator        pKind     pH range

litmus                  6.5       5-8
methyl orange       3.7     3.1 - 4.4

phenolphthalein     9.3    8.3 - 10.0

The litmus color change happens over an unusually wide range, but it
is useful for detecting acids and alkalis in the lab because it changes
color around pH 7. Methyl orange or phenolphthalein would be less
useful.

This is more easily seen diagramatically.

For example, methyl orange would be yellow in any solution with a pH
greater than 4.4. It couldn't distinguish between a weak acid with a pH
of 5 or a strong alkali with a pH of 14.

Choosing indicators for titrations

Remember that the equivalence point of a titration is where you have
mixed the two substances in exactly equation proportions. You
obviously need to choose an indicator which changes color as close
as possible to that equivalence point. That varies from titration to
titration.

Strong acid v strong base

The next diagram shows the pH curve for adding a strong acid to a
strong base. Superimposed on it are the pH ranges for methyl orange
and phenolphthalein.
You can see that neither indicator changes color at the equivalence
point.

However, the graph is so steep at that point that there will be virtually
no difference in the volume of acid added whichever indicator you
choose. However, it would make sense to titrate to the best possible
color with each indicator.

If you use phenolphthalein, you would titrate until it just becomes
colorless (at pH 8.3) because that is as close as you can get to the
equivalence point.

On the other hand, using methyl orange, you would titrate until there is
the very first trace of orange in the solution. If the solution becomes
red, you are getting further from the equivalence point.

Strong acid v weak base
This time it is obvious that phenolphthalein would be completely
useless. However, methyl orange starts to change from yellow
towards orange very close to the equivalence point.

You have to choose an indicator which changes color on the steep bit
of the curve.

Weak acid vs strong base

This time, the methyl orange is hopeless! However, the
phenolphthalein changes color exactly where you want it to.

Weak acid v weak base
The curve is for a case where the acid and base are both equally
weak - for example, ethanoic acid and ammonia solution. In other
cases, the equivalence point will be at some other pH.

You can see that neither indicator is any use. Phenolphthalein will
have finished changing well before the equivalence point, and methyl
orange falls off the graph altogether.

It may be possible to find an indicator which starts to change or
finishes changing at the equivalence point, but because the pH of the
equivalence point will be different from case to case, you can't
generalise.

On the whole, you would never titrate a weak acid and a weak base in
the presence of an indicator.

Sodium carbonate solution and dilute hydrochloric acid

This is an interesting special case. If you use phenolphthalein or
methyl orange, both will give a valid titration result - but the value with
phenolphthalein will be exactly half the methyl orange one.
It so happens that the phenolphthalein has finished its color change at
exactly the pH of the equivalence point of the first half of the reaction
in which sodium hydrogencarbonate is produced.

The methyl orange changes color at exactly the pH of the equivalence
point of the second stage of the reaction.

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