Chapter 17 : Additional Aspects of Aqueous Equilibria

Chapter 17: Additional Aspects of Aqueous Equilibria

Section 2: Buffered Solutions

Introduction

Solutions prepared with common ions have a tendency to resist drastic pH changes even when subjected to the addition of strong acids or bases.

Such solutions are called buffered solutions.

Blood is buffered to a pH of about 7.4.

Surface seawater is buffered to a pH between 8.1 and 8.3.

Composition and Action

A buffer resists pH change because it contains … an acid to neutralize added OH− ions a base to neutralize added H+ ions

But we cannot have the acid and base consume one another through neutralization.


Therefore, we use weak conjugate acids and bases to prepare our buffers.

For example: HCH3COO and CH3COO−

HF and F−

NH4+ and NH3

HClO and ClO−


If we have a solution of HF and NaF, we have HF(aq) and F−(aq) in solution.

If we add a small amount of a strong acid, we use the F− to neutralize it:

F−(aq) + H+(aq) → HF(aq) If we add a small amount of a strong

base, we use the HF to neutralize it:

HF (aq) + OH−(aq) → F−(aq) + H2O(l)


Let’s consider another buffer composed of the weak acid HX and the strong electrolyte MX (M is a metal from Group 1, for example).

The acid dissociation involves both the weak acid and its conjugate base.

HX(aq) ⇄ H+(aq) + X−(aq)




HX(aq) ⇄ H+(aq) + X−(aq) The acid dissociation constant is

Ka =

[H+][X−][HX]




HX(aq) ⇄ H+(aq) + X−(aq) Solving for [H+]

[H+] = Ka

[HX][X−]


[H+] = Ka[HX][X−]


[H+] = Ka

The [H+] and pH depend on two things:

[HX][X−]


[H+] = Ka

The [H+] and pH depend on two things: Ka

the ratio of the concentrations of conjugate acid-base pair.

[HX][X−]


[H+] = Ka

If OH− ions are added, they will react with the acid component to produce water and X−.

[HX][X−]


[H+] = Ka


OH−(aq) + HX(aq) → H2O(l) + X−(aq)

[HX][X−]


[H+] = Ka


OH−(aq) + HX(aq) → H2O(l) + X−(aq) This causes a decrease in [HX] and

an increase in [X−].

[HX][X−]


[H+] = Ka


OH−(aq) + HX(aq) → H2O(l) + X−(aq) This causes a decrease in [HX] and an

increase in [X−]. So long as the amounts of HX and X− are

large, the ratio of [HX]/[X−] does not change.

[HX][X−]


[H+] = Ka

If H+ ions are added, they will react with the base component to produce the acid.

[HX][X−]


[H+] = Ka


H+(aq) + X−(aq) → HX (aq)

[HX][X−]


[H+] = Ka


H+(aq) + X−(aq) → HX (aq) This causes a decrease in [X−] and

an increase in [HX].

[HX][X−]


[H+] = Ka


H+(aq) + X−(aq) → HX (aq) This causes a decrease in [X−] and an

increase in [HX]. So long as the amounts of HX and X− are

large, the ratio of [HX]/[X−] does not change.

[HX][X−]

Calculating the pH of a Buffer

We could use the technique of Sample Problem 17.1 to calculate the pH of a buffer.



But, there is an alternative method.



But, there is an alternative method. We start with the expression for

finding [H+] of a buffer.



But, there is an alternative method. We start with the expression for

finding [H+] of a buffer.[HX][X−][H+] =

Ka



But, there is an alternative method. We then find the negative log of both

sides.[HX][X−][H+] =

Ka



sides.−log[H+] = −log(Ka

)


[HX][X−]



sides.−log[H+] = −logKa

− log


[HX][X−]


But, there is an alternative method. We note that −log[H+] = pH and

−logKa = pKa

−log[H+] = −logKa − log


[HX][X−]


But, there is an alternative method. We note that −log[H+] = pH and

−logKa = pKa

pH = pKa − log


[HX][X−]


But, there is an alternative method. We note that − log[HX]/[X−] = +

log[X−]/[HX]pH= pKa − log


[HX][X−]


But, there is an alternative method. We note that − log[HX]/[X−] = +

log[X−]/[HX]pH= pKa + log


[X−][HX]


But, there is an alternative method. We can generalize HX as an acid and

X− as a base.pH= pKa + log


[X−][HX]


But, there is an alternative method. We can generalize HX as an acid and

X− as a base.pH= pKa + log


[base][acid]


But, there is an alternative method. This is called the Henderson-

Hasselbach equation.pH= pKa + log


[base][acid]

What is the pH of a buffer that is 0.12 M in lactic acid, CH3CH(OH)COOH, and 0.10 M in sodium lactate, NaCH3CH(OH)COO? For lactic acid, Ka = 1.4 × 10−4.

Sample Exercise 17.3 (pg. 725)


Known:



Known: Ka = 1.4 × 10−4



Known: Ka = 1.4 × 10−4 ⇒ pKa = −logKa



Known: Ka = 1.4 × 10−4 ⇒ pKa = −logKa = 3.85




[base] = 0.10 M




[base] = 0.10 M[acid] = 0.12 M




[base] = 0.10 M[acid] = 0.12 M

pH = pKa + log


[base][acid]



[base] = 0.10 M[acid] = 0.12 M

pH = 3.85+ log


0.100.12



[base] = 0.10 M[acid] = 0.12 M

pH = 3.85+ log = 3.85 + log(0.83)


0.100.12



[base] = 0.10 M[acid] = 0.12 M

pH = 3.85+ log = 3.85 − 0.08


0.100.12



[base] = 0.10 M[acid] = 0.12 M

pH = 3.85+ log = 3.85 − 0.08 = 3.77


0.100.12

Use Sample Exercise 17.3 to help solve homework exercises 17.21 through 17.24 (pages 759 − 760).


How many moles of NH4Cl must be added to 2.0 L of 0.10 M NH3 to form a buffer whose pH = 9.00? (Assume no volume change when adding the NH4Cl.)



Known:



Known: Kb = 1.8 × 10−5



Known: Kb = 1.8 × 10−5 ⇒



Known: Kb = 1.8 × 10−5 ⇒ Ka = 5.6 × 10−10



Known: Kb = 1.8 × 10−5 ⇒ Ka = 5.6 × 10−10

pH = 9.00



Known: Kb = 1.8 × 10−5 ⇒ Ka = 5.6 × 10−10

pH = 9.00 ⇒



Known: Kb = 1.8 × 10−5 ⇒ Ka = 5.6 × 10−10

pH = 9.00 ⇒ [H+] = 1.0 × 10−9



Known: Kb = 1.8 × 10−5 ⇒ Ka = 5.6 × 10−10

pH = 9.00 ⇒ [H+] = 1.0 × 10−9

[base] = 0.10 M



Known: Kb = 1.8 × 10−5 ⇒ Ka = 5.6 × 10−10

pH = 9.00 ⇒ [H+] = 1.0 × 10−9

[base] = 0.10 M

[H+] = Ka


[acid][base]


Known: Kb = 1.8 × 10−5 ⇒ Ka = 5.6 × 10−10

pH = 9.00 ⇒ [H+] = 1.0 × 10−9

[base] = 0.10 M

[H+] = Ka ⇒ [acid] =


[acid][base]

[H+][base]Ka


Known: Kb = 1.8 × 10−5 ⇒ Ka = 5.6 × 10−10

pH = 9.00 ⇒ [H+] = 1.0 × 10−9

[base] = 0.10 M

[acid] =


[H+][base]Ka


Known: Kb = 1.8 × 10−5 ⇒ Ka = 5.6 × 10−10

pH = 9.00 ⇒ [H+] = 1.0 × 10−9

[base] = 0.10 M

[acid] = =


[H+][base]Ka

(1.0 × 10−9)(0.10)5.6 × 10−10


Known: Kb = 1.8 × 10−5 ⇒ Ka = 5.6 × 10−10

pH = 9.00 ⇒ [H+] = 1.0 × 10−9

[base] = 0.10 M

[acid] = = 0.18 M


[H+][base]Ka


Known: Kb = 1.8 × 10−5 ⇒ Ka = 5.6 × 10−10

pH = 9.00 ⇒ [H+] = 1.0 × 10−9

[base] = 0.10 M

[acid] = = 0.18 M = [NH4+]


[H+][base]Ka


Known: Kb = 1.8 × 10−5 ⇒ Ka = 5.6 × 10−10

pH = 9.00 ⇒ [H+] = 1.0 × 10−9

[base] = 0.10 M

[NH4+] = 0.18 M



Known: Kb = 1.8 × 10−5 ⇒ Ka = 5.6 × 10−10

pH = 9.00 ⇒ [H+] = 1.0 × 10−9

[base] = 0.10 M

[NH4+] = 0.18 M ⇒



Known: Kb = 1.8 × 10−5 ⇒ Ka = 5.6 × 10−10

pH = 9.00 ⇒ [H+] = 1.0 × 10−9

[base] = 0.10 M

[NH4+] = 0.18 M ⇒ n = M × V



Known: Kb = 1.8 × 10−5 ⇒ Ka = 5.6 × 10−10

pH = 9.00 ⇒ [H+] = 1.0 × 10−9

[base] = 0.10 M

[NH4+] = 0.18 M ⇒ n = (0.18 M)(2.0 L)



Known: Kb = 1.8 × 10−5 ⇒ Ka = 5.6 × 10−10

pH = 9.00 ⇒ [H+] = 1.0 × 10−9

[base] = 0.10 M

[NH4+] = 0.18 M ⇒ n = 0.36 moles


Use Sample Exercise 17.4 to help solve homework exercises 17.25 through 17.26 (page 760).


There are two very important characteristics of buffers. the buffer capacity the effective pH range

Buffer Capacity and pH Range

The buffer capacity is the amount of acid or base a buffer can neutralize before the pH begins to change to an appreciable degree.

Buffer capacity depends on … the amount of buffer materials the identity of the buffer materials


The pH range of any buffer is the range of pH over which a buffer works to effectively neutralize an acid or a base.

The optimal situation is where the concentration of the buffer acid is equal to the concentration of the buffer base.

In that case, pH = pKa. Buffers usually have a pH range of about

±1 pH unit of pKa.


Next, we’ll look at what happens when we add a strong acid or strong base to a buffer.

This is going to be a quantitative approach (that means numbers!). Remember …▪ Strong acids react with weak bases to completion.▪ H+ + ClO− → HClO

▪ Strong bases react with weak acids to completion.▪ OH− + CH3COOH → H2O + CH3COO−

Addition of Strong Acid or Base

To calculate how the pH of a buffer changes when adding a strong acid or base, we follow the following strategy. First, consider the acid-base

neutralization reaction and determine its effect on [HX] and [X−].

Use the Ka and new values for [HX] and [X−] to calculate [H+].▪ Use i-c-e table or Henderson-Hasselbach

(easiest).

Addition of Strong Acid or Base

A buffer is made by adding 0.300 mol CH3COOH and 0.300 mol CH3COONa to enough water to make 1.00 L of solution.

Sample Exercise 17.5 (page 728)

A buffer is made by adding 0.300 mol CH3COOH and 0.300 mol CH3COONa to enough water to make 1.00 L of solution. The pH of the buffer is 4.74 (Sample Exercise 17.1).



a) Calculate the pH of this solution after 0.020 mol NaOH is added.




b) For comparison, calculate the pH that would result if 0.020 mol NaOH were added to pure water.



First we need to do the neutralization.




OH− + CH3COOH → CH3COO− + H2O






ninitial 0.020 0.300 0.300





ninitial 0.020 0.300 0.300

nchg −0.020 −0.020 +0.020





ninitial 0.020 0.300 0.300

nchg −0.020 −0.020 +0.020

nfinal 0.000 0.280 0.320


First we need to do the neutralization.OH− + CH3COOH → CH3COO− +

H2O

This give us: [CH3COOH] = 0.280 M


ninitial 0.020 0.300 0.300

nchg −0.020 −0.020 +0.020

nfinal 0.000 0.280 0.320



H2O

This give us: [CH3COOH] = 0.280 M[CH3COO−] = 0.320 M


ninitial 0.020 0.300 0.300

nchg −0.020 −0.020 +0.020

nfinal 0.000 0.280 0.320



H2O














Using Henderson-Hasselbach:






pH = pKa + log


[base][base]





pH = 4.74+ log


[0.280]

[0.320]





pH = 4.74+ log(1.14)






pH = 4.74+ log(1.14) = 4.74 + 0.06



First we need to do the neutralization.This give us: [CH3COOH] = 0.280 M

[CH3COO−] = 0.320 M Using Henderson-Hasselbach:

pH = 4.74+ log(1.14) = 4.74 + 0.06 = 4.80





In pure water, NaOH is fully dissociated.



In pure water, NaOH is fully dissociated. Therefore, [OH−] = 0.020 mol/1.00 L



In pure water, NaOH is fully dissociated. Therefore, [OH−] = 0.020 mol/1.00 L = 0.020 M.




pOH = −log[OH−]




pOH = −log[OH−] = −log(0.020)




pOH = −log[OH−] = −log(0.020) = 1.70




pOH = −log[OH−] = −log(0.020) = 1.70pH = 14.00 − pOH




pOH = −log[OH−] = −log(0.020) = 1.70pH = 14.00 − pOH = 14.00 − 1.70




pOH = −log[OH−] = −log(0.020) = 1.70pH = 14.00 − pOH = 14.00 − 1.70 = 12.30


Documents

Chapter 17 : Additional Aspects of Aqueous Equilibria