Question

(a) Calculate the pHin a solution prepared by dissolving 0.050molof acetic acid and 0.020molof sodium acetate in water and adjusting the volume to.

(b) Suppose 0.010molofNaOH is added to the buffer from part (a). Calculate the pHof the solution that results.

Short answer

$\begin{array}{l}a)pH=4.36\\ \text{b)pH}=4.63\end{array}$

Step-by-step solution

Formula used

$c\left(\text{concentration}\right)=\frac{\text{n(numberofmoles})}{\text{V}\left(\text{volume}\right)}$

Formula to calculate pH of the solution:

$\text{pH}=-\log \left(\left[{\text{H}}^{+}\right]\right)$

Given information from question

a)

$\begin{array}{l}n\left({\text{CH}}_3\text{COOH}\right)=0.050\text{mol}\\ n\left({\text{CH}}_3\text{COONa}\right)=0.020\text{mol}\\ \text{V}=500\text{mL}=0.5\text{L}\end{array}$

Calculate the concentration of acetic acid and acetate ion

The concentration is

${\text{CH}}_{\text{3}}\text{COONa}\rightleftharpoons {\text{CH}}_{\text{3}}{\text{COO}}^{\text{-}}{\text{+Na}}^{\text{+}}$

$\begin{array}{l}\text{c}\left({\text{CH}}_3\text{COOH}\right)=\frac{0.050\text{mol}}{0.5\text{L}}\\ \text{c}\left({\text{CH}}_3\text{COOH}\right)=0.10\text{M}\end{array}$

Do the same for acetate ion:

$\begin{array}{l}\text{c}\left({\text{CH}}_3{\text{COO}}^{-}\right)=\frac{\text{n}\left({\text{CH}}_{3}3{\text{COO}}^{-}\right)}{\text{V}}\\ \text{c}\left({\text{CH}}_{3}CO{O}^{-}\right)=\frac{0.020\text{mol}}{0.5\text{L}}\\ =0.04M\end{array}$

Write equilibrium table

Equilibrium for the given buffer is

Concentration(M)	$C{H}_{3}COOH\left(aq\right)$		H+(aq)	CH3COO-(aq)
Initial	0.10		0	0.04
Change	-X		+X	+X
equilibrium	0.10-X		+X	0.04+4

Write equilibrium constant

The formula of equilibrium constant is

$\text{K}a=\frac{\left[{\text{H}}^{+}\right]\left[\text{CH}3{\text{COO}}^{-}\right]}{\left[\text{CH}3\text{COOH}\right]}$

Put the data from the equilibrium constant table knowing that the equilibrium constant for acetic acid is $1.76\times {10}^{-5}$

$\begin{array}{l}1.76\times {10}^{-5}=\frac{x\times (0.04+x)}{0.10-x}\\ 1.76\times {10}^{-6}-1.76\times {10}^{-5}x=0.04x+x^2\\ 1.76\times {10}^{-6}-1.76\times {10}^{-5}\text{x}-0.04\text{x}-{\text{x}}^2=0\\ {\text{x}}^2+0.04\text{x}-1.76\times {10}^{-6}=0\end{array}$

Solve this quadratic equation to get which is the $\left[{\text{H}}^{+}\right]$vale:

$x=4.36\times {10}^{-5}$

 Step 6: Calculate pH

Put data in the formula to calculate pH:

$\begin{array}{l}\text{pH}=-\log \left(\left[{\text{H}}^{+}\right]\right)\\ \text{pH}=-\log (4.36\times {10}^{-5})\\ pH=4.36\end{array}$

Given information from question

b)$\text{n}\left(\text{NaOH}\right)=0.010\text{mol}$

Calculate the concentration

Put the data for the number of moles and volume in the formula:

$\begin{array}{l}c\left(\text{NaOH}\right)=\frac{n\left(\text{NaOH}\right)}{\text{V}}\\ c\left(\text{NaOH}\right)=\frac{0.010\text{mol}}{0.5\text{L}}\\ \text{c}\left(\text{NaOH}\right)=0.02\text{M}\end{array}$

The equilibrium for NaOH and CH₃COOH is:

$\text{NaOH}+\text{CH}3\text{COOH}\rightleftharpoons \text{CH}3\text{COONa}+\text{H}2\text{O}$

We are asked to determine the of a buffer after addition of a small amount of strong base. We have to do a stoichiometry calculation to determine how the added OH^- reacts with the buffer and affects its composition.

Table to see how the composition of the buffer changes as a result of its reaction with OH

The table is:

Concentration(M)	CH3COOH(aq)	OH-(aq)		H2O(l)		CH3COO-(aq)
Buffer before addition	0.10	0				0.04
Addition	-X	-0.02				+X
Buffer after addition	0.08	0				0.06

Calculate the pH

The using formula:

$\begin{array}{l}\text{pH}=\text{pK}a-\log \frac{\left[\text{CH}3\text{COOH}\right]}{\left[\text{CH}3{\text{COO}}^{-}\right]}\\ pKais4.76,\\ \text{pH}=4.76-\log \left(\frac{0.08\text{M}}{0.06\text{M}}\right)\\ \text{pH}=4.63\end{array}$