Question

Suppose there is 1 .00 L of an aqueous buffer containing 60.0 mmol of acetic acid (pKa = 4.76) and 40.0 mmol of acetate. Calculate the pH of this buffer pH= | 4.58 What volume of 2.00 M NaOH would be required to increase the pH to 4.93? TOOLS x10

Answer 1

Volume of the solution is 1.00 L

Initial concentration of acetic acid and acetate ions are

$[CH_3COOH] = 60.0 \ mmol/L = 6.00\times 10^{-2} \ M$

$[CH_3COO^-] = 40.0 \ mmol/L = 4.00\times 10^{-2} \ M$

The concerned dissociation of the acid can be written as

$CH_3COOH_{(aq)} + H_2O_{(l)} \rightarrow CH_3COO^-_{(aq)} + H_3O^+_{(aq)}$

Hence, we can create an ICE table to determine the equilibrium concentrations

	$[CH_3COOH]$	$[CH_3COO^-]$	$[H_3O^+]$
Initial, M	0.0600	0.0400	0
Change, M	-x	+x	+x
Equilibrium, M	0.0600 -x	0.0400+x	x

Given that the pKa of the acid is 4.76

$K_a - 10^{-pK_a} = 10^{-4.76} = 1.74 \times 10^{-5}$

hence,

$K_a = \frac{[CH_3COO^-]\times [H_3O^+]}{[CH_3COOH]} = \frac{(0.04+x)\times x }{0.06+x} = 1.74 \times 10^{-5} \\ \Rightarrow x = 2.61 \times 10^{-5} \ M$

Hence,

$pH = -\log[H_3O^+] = -\log x = -\log (2.61\times 10^{-5}) = +4.58$

Hence, the pH of the solution is 4.58.

Alternatively, you can directly use Henderson-Hasselbalch equation

$pH = pK_a + \log \frac{[base]}{[acid]} = 4.76 + \log \frac{0.04}{0.06} = 4.76 - 0.176 \approx 4.58$

We have to increase the pH by adding NaOH in the solution. Addition of NaOH can be represented by

$CH_3COOH_{aq} + NaOH _{aq} \rightarrow H_2O _{l}+ CH_3COO^-_{aq}$

Hence, the ratio of base to acid can be calculated by setting the target pH as 4.93

$4.93 = 4.76 + \log \frac{[base]}{[acid]}\\ \Rightarrow \log \frac{[base]}{[acid]} = 0.17 \\ \Rightarrow \frac{[base]}{[acid]} = 10^{0.17} \approx 1.48$

Note that addition of NaOH results in increase in the concentration of CH3COO- base and decrease in concentration of acid CH3COOH as the equation above suggest.

Hence, we can create a new ICE table

	$[CH_3COOH]$	$[OH^-]$	$[CH_3COO^-]$
Initial, mol	0.0600 -x	y	0.0400+x
Change, mol	-y	-y	+y
Equilibrium, mol	0.06-x-y	0	0.0400+x+y

Note that we have to add y mol of NaOH which would completely react with the weak acid and produce conjugate base equal to the same mol y.

Now since x is very less that 0.06 or 0.04, we can neglect it

Hence, the new equilibrium moles of acid and base are

$[CH_3COOH] = 0.06 -y$

$[CH_3COO^-] = 0.04 +y$

Hence, using the calculated ratio of acid and base above

$\frac{[base]}{[acid]} = \frac{[CH_3COO^-]}{[CH_3COOH]} = \frac{0.04+y}{0.06-y} = 1.48 \\ \Rightarrow y = 0.0197$

Hence, we need to add 0.0197 mol of NaOH to increase the pH to 4.93.

The concentration of NaOH solution available is 2.00 M = 2.00 mol/L

Hence, the volume that contains 0.0197 mol NaOH is

$\frac{1000 \ mL}{2.00 \ mol} \times 0.0197 \ mol \approx 9.85 \ mL$

Hence, the required volume of NaOH to be added is approximately 9.85 mL.