Question

A buffer solution is prepared from equal volumes of \({\bf{0}}.{\bf{200}}{\rm{ }}{\bf{M}}\) acetic acid and \({\bf{0}}.{\bf{600}}{\rm{ }}{\bf{M}}\) sodium acetate. Use \({\bf{1}}.{\bf{80}}{\rm{ }} \times {\rm{ }}{\bf{1}}{{\bf{0}}^{ - {\bf{5}}}}\) as K_a for acetic acid.

(a) What is the pH of the solution?

(b) Is the solution acidic or basic?

(c) What is the pH of a solution that results when \({\bf{3}}.{\bf{00}}{\rm{ }}{\bf{mL}}\) of \({\bf{0}}.{\bf{034}}{\rm{ }}{\bf{M}}{\rm{ }}{\bf{HCl}}\) is added to \({\bf{0}}.{\bf{200}}{\rm{ }}{\bf{L}}\) of the original buffer?

Short answer

a) pH = 5.22

b) The solution is acidic.

c) pH = 5.22

Step-by-step solution

pH of solution

a) In order to calculate the pH of a buffer solution, we use the Henderson-Hasselbach equation

\(\begin{align}pH &= p{K_a} + log\left( {\frac{{c( salt )}}{{c( acid )}}} \right)\\pH &= - log\left( {1.8 \times 1{0^{ - 5}}} \right) + log\left( {\frac{{0.600}}{{0.200}}} \right)\\pH &= 5.22\end{align}\)

acid or base?

b) The solution is acidic because its pH is below 7 (5.22).

pH of solution after adding HCl

c) In order to calculate the pH change, we need to assume that the only thing that happens is that the \({H^ + }\)ions react with \(C{H_3}CO_2^ - \)ions and form \(C{H_3}C{O_2}H\)molecules. The amount of \({H^ + }\)ions added is:

\(\begin{align}n &= c \times V\\n &= 0.034\frac{{mol}}{L} \times 0.003L\\n &= 1.02 \times 1{0^{ - 4}}\;mol\end{align}\)

In the HH equation, we can write the n of the acid and the base instead of the c because

\(c = \frac{n}{V}\)

and since the volumes are the same, they cancel each other out, therefore:

\(pH = p{K_a} + log\left( {\frac{{n( salt )}}{{n(acid)}}} \right)\)

Since we only use 0.200 L of the buffer, which is\(\frac{{1.00L}}{5}\), we can divide the

concentrations by 5 to get the amount of the acid and the base component :

\(\begin{align}n( salt ) &= \frac{{0.600}}{5}\\n( salt ) &= 0.12\;mol\\n( acid ) &= \frac{{0.200}}{5}\\n( acid ) &= 0.04\;mol\end{align}\)

Since \({H^ + }\)ions react with \(C{H_3}CO_2^ - \)ions, their amount drops by \(1.02 \times 1{0^{ - 4}}\;mol\)while the amount of \(C{H_3}C{O_2}H molecules rises by 1.02 \times 1{0^{ - 4}} mol\)after the addition of HCl.

\(\begin{align}pH &= p{K_a} + log\left( {\frac{{n(salt)}}{{n(acid)}}} \right)\\pH &= - log\left( {1.8 \times 1{0^{ - 5}}} \right) + log\left( {\frac{{0.12 - 1.02 \times 1{0^{ - 4}}}}{{0.04 + 1.02 \times 1{0^{ - 4}}}}} \right)\\pH &= 5.22\end{align}\)

If we add the written amount of HCl, the pH of the buffer solution remains practically unchanged at 5.22.

Final answer

a) pH=5.22

b) The solution is acidic.

c) pH=5.22