All About Buffers

Chemistry 142 B

Autumn Quarter 2004

J. B. Callis, Instructor

Lecture #23

Solutions of Acids or Bases Containing a Common Ion

• Previously, in dealing with a solution of a weak acid HA we have assumed that the salt of the acid NaA is not initially present.

• Now we assume that both the acid and its salt are present.

• The effect of the simultaneous presence of the salt is to lower the extent of dissociation of the acid.

• This is called the common ion effect. • The common ion effect occurs with polyprotic

acids and mono- and polyprotic bases as well.

Problem 23-1: pH of a Weak Acid With and Without a Common Ion

Acetic acid (HAc) is a weak acid with Ka = 1.8 x 10-5. Calculate the pH and percent dissociation of acetic acid in a solution containing 0.1 M acetic acid and 0.05 M sodium acetate (NaAc).

Ans: This is a sufficiently concentrated solution that the autoionization of water can be neglected.

The major species in solution are:

The relevant equilibrium is

Problem 23-1: Continued (a) Construct the reaction table:

Conc. (M)

HAc H+(aq) Ac-(g)

Init.

Change

Equil.

Problem 23-1: Continued (b)

%100HAc

H

ison dissociatipercent The

pH H

:root positive thekeeping and quadratic thisSolving

108.1

0

5

x

HAc

AcHK

Problem 23-1: Continued (c)

Compare these values of [H+] and percent dissociation of HAc with those for an 0.1 M HAc solution, where [H+] = 1.3 x 10-3 M and the percent dissociation is 1.3%.

Ans:

Buffered Solutions

• A buffered solution is one that resists changing pH when acid or bases is added.

• A buffered solution contains a weak acid and its salt or a weak base and its salt.

• By choosing appropriate components buffering can be had at virtually any pH.

• The calculations for a buffer solution are similar to those for the common ion effect.

Problem 23-2: pH of a Buffer Solution Before and After Addition of Strong Base

Hydrofluoric acid (HF) is a weak acid with Ka = 6.8 x 10-4. Calculate the pH and percent dissociation of HF in a solution containing 0.5 M HF and 0.45 M sodium fluoride (NaF) (a) before, and (b) after addition of 0.40 g NaOH to 1.0 L of the buffer.

Ans:

The major species in solution are:

The relevant equilibrium is

Problem 23-2(a)- Continued (i) Construct the reaction table:

Conc. (M)

HF(aq) H+(aq) F-(aq)

Init.

Change

Equil.

Problem 23-2(a): Continued (ii)

pH ; H

:root positive thekeeping and quadratic thissolving

108.6 4

x

HF

FHK

Problem 23-2(b): pH of a Buffer Solution After Addition of Strong Base

What happens after addition of 0.40 g NaOH to 1.0 L of the buffer of 2(a)?

Ans: The major species in solution are:

The reaction that occurs is:

First assume the reaction goes to completion, then let the resultant species establish a new equilibrium according to:

Problem 23-2(b)- Continued (i) After 0.5 M HF reacts with 0.01 M NaOH we can construct the reaction table:

Conc. (M)

HF(aq) H+(aq) F-(aq)

Init.

Change

Equil.

Problem 23-2(b): Continued (ii)

pH ;M H

:root positive thekeeping and quadratic thissolving

108.6 4

x

HF

FHK

Compare a Buffer to an Unbuffered Solution

23-2(c): Suppose that the same .01 mol NaOH is added to 1.0 L of water to give .01 M NaOH. In this case, [OH-] = 0.01 M and

is pHin change theThus

pH

OH

KH

w

How Does a Buffer Work?Start by considering what happens to a buffered solution when OH- is added. The following reaction occurs:

OH-(aq) + HA(aq) = A-(aq) + H2O(l)

Thus, OH- does not build up, instead A- increases, but there is already a lot of A- already.

The stability of pH under these conditions can be seen from the

equilibrium expression:

A

HAKH g,rearranginor

HA

AHaaK

Clearly, the pH is determined by the ratio [HA]/[A-]

Henderson-Hasselbalch Equation

acid

baselogpK

HA

AlogpKpH aa

For a particular buffering solution (acid-conjugate base pair), all solutions that have the same ratio [A-]/[HA] have the same pH.

When HA and A- are the starting concentrations, the Henderson – Hasselbalch equation is an approximation, i.e. it requires

HAHAHA andAAA 0000 xx

Problem 23-3: Buffer Calculation Using the Henderson-Hasselblach Equation

A buffered solution contains 0.35 M of CH3NH2 (Kb = 4.38 x 10-4 and 0.55 M CH3NH3

+

a. Calculate the pH of this solution.

x

NHCH

OHNHCHKb

23

33

The approximations are valid by the 5% rule, so

[H+] =

pH =

Problem 23-3(cont.) - part b

b. Now calculate the pH of this solution using the Henderson- Hasselbalch approach:

H

logKH

p

acid

basepp a

Problem 3(cont.) - part c

b. Calculate the pH of the solution that results when 0.10 M gaseous HCl is added to the buffered solution from part a.

Ans: assume that the reaction goes essentially to completion. So [acid]=0.65 M and [base]=0.25 M

H

logKH

p

acid

basepp a

Exact Treatment of Buffered Solutions• We have the following four unknowns to solve

for: [H+], [OH-], [HA] and [A-]• We have the following four equations which

govern the concentrations of the above species:

balance material AHAAHA

balance charge OHAAH

hydrolysister wa OHHK

mequilibriu base-acid HA

AHK

0-

0

0

w

a

The Exact Solution to the Buffer Problem

We now proceed to solve the four equations for [H+] by successive elimination of variables. The result for [H+] is

[H+]3 +(Ka+[A-]0)[H+]2 – (Kw +Ka[HA]0)[H+] –KaKw = 0

Finding [H+] requires that we find the three roots of the above cubic equation and then selecting the one root that is consistent with physical reality, i.e. leads to all positive concentrations. Fortunately, in the above case only one root is positive and the other two are negative.

This positive root can be found by the ‘Solver’ function of your calculator. You can make a wild guess that is positive and the calculator will converge to the correct answer.

Problem 23-4: pH of a Very Weak Buffer

Find the pH of a solution of 1.0 x 10-7 M vitamin B1, thiamine hydrochloride (HC12H17ON4SCl2, mw = 337.27 g/mol ), a weak acid with Ka = 3.4 x 10-7 and its conjugate base at a concentration of 2 x 10-7.

Problem 23-4: pH of a Very Weak Buffer – cont. (a)

Start with the expression for the value of [H+] for a buffer solution as derived for the case where the autoionization of water cannot be neglected:

Now substitute the relevant values for the constants of the problem, Ka, [HA]0 and Kw:

Where y = [H+]

This expression may be simplified to:

Problem 23-4: pH of a Very Weak Buffer – cont. (b)

To solve:

y3 + (5.4 x 10-7)y2 – (4.4 x 10-14)y – 3.4 x 10-21 = 0

We need to find the physically relevant root. This may be obtained using the ‘Solver’ function of the TI-83 calculator.

Solver requires a guess for the answer and boundaries.

We know that the desired root must be greater than zero and less than 3.4 x 10-7 M. So these can be used as the boundaries. We know that the pH will be near neutral and we have an acid, so we try 10-7 as the guess for y.

We punch in the values and BAM! The answer comes up:

[H+] = ; pH =

Problem 23-4: Neglect the Water IonizationLet’s attempt the problem ignoring the water ionization

pHx

acid

HbaseK a

;

0

104.3 7

This pH is on the side of 7, which is impossible when an acid is dissolved in water.

Buffer Capacity

• Buffer capacity is a measure of the amount of protons or hydroxide ions a solution can absorb without a significant change in pH.

• A buffer with a large capacity contains large amounts of buffering components.

• While the pH of a buffer is determined by the ratio of base to acid, the capacity of a buffer is determined by magnitudes of concentrations of acid and base.

Answers to Problems in Lecture 231. (a) pH=4.44; (b) 0.036%; (c) The large difference

clearly shows the presence of the Ac- ions from the dissolved NaAc greatly inhibits the dissociation of HAc. The position of the acid dissociation equilibrium has been shifted to the left by the presence of Ac- ions from NaAc.

2. (a) pH = 3.12; (b) pH = 3.14; (c) pH increases by +5 units.

3. (a) pH = 10.44; (b) pH = 10.44; (c) pH = 10.23

4. pH = 6.95; ignoring water ionization, pH = 7.22 (incorrect).