A \({\bf{5}}{\bf{.36 g}}\) sample of \({\bf{N}}{{\bf{H}}_{\bf{4}}}{\bf{Cl}}\) was added to \({\bf{25}}.{\bf{0}}{\rm{ }}{\bf{mL}}\) of \({\bf{1}}{\bf{.00 M NaOH}}\) and the resulting solution diluted to\({\bf{0}}.{\bf{100}}{\rm{ }}{\bf{L}}\).

(a) What is the pH of this buffer 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 the solution?

a) There must be a mistake in the problem text because a mixture of \(N{H_4}Cl\)and \(NaOH\)is not a buffer solution. However, a mixture of \(N{H_4}Cl\) and \(N{H_3}\)is a buffer solution and we will therefore assume that it was supposed to write \(N{H_3}\)instead of\(NaOH\). In order to calculate the pH, we can use the HH equation modified for a "base-buffer":

\(\begin{align}{\rm{N}}{{\rm{H}}_3}(aq) + {{\rm{H}}_2}{\rm{O}}({\rm{l}}) \to {\rm{NH}}_4^ + (aq) + {\rm{O}}{{\rm{H}}^ - }(aq)\\{K_b} = \frac{{c\left( {{\rm{NH}}_4^ + } \right) \cdot c\left( {O{H^ - }} \right)}}{{c\left( {N{H_3}} \right)}}\end{align}\)

Now we multiply the whole equation with\( - log\):

\(\begin{align}p{K_b} &= pOH - log\left( {\frac{{c\left( {NH_4^ + } \right)}}{{c\left( {N{H_3}} \right)}}} \right)\\pOH &= p{K_b} + log\left( {\frac{{c\left( {NH_4^ + } \right)}}{{c\left( {N{H_3}} \right)}}} \right)\end{align}\)

We can look up the \({K_b}\)value for \(N{H_3}\)in the table in Appendix I in the book. In the HHequation, 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:

\(\begin{align}pOH &= p{K_b} + log\left( {\frac{{n\left( {NH_4^ + } \right)}}{{n\left( {N{H_3}} \right)}}} \right)\\n\left( {NH_4^ + } \right) &= \frac{m}{M}\end{align}\)

M is the molar mass, and it is calculated as the sum of all the relative atomic masses of the elements composing the molecule. We can look up the relative atomic masses in any periodic table of elements.

\(\begin{align}M\left( {N{H_4}Cl} \right) &= (Ar(N) + 4Ar(H) + Ar(Cl))\frac{g}{{mol}}\\M\left( {N{H_4}Cl} \right) &= (14.01 + 4 \times 1.01 + 35.45)\frac{g}{{mol}}\\M\left( {N{H_4}Cl} \right) &= 53.50\frac{g}{{mol}}\end{align}\)

\(\begin{align}n\left( {NH_4^ + } \right) &= \frac{m}{M}\\n\left( {NH_4^ + } \right) &= \frac{{5.36g}}{{53.50{\rm{gmo}}{{\rm{l}}^{ - 1}}}}\\n\left( {{\rm{NH}}_4^ + } \right) &= 0.10\;{\rm{mol}}\\n\left( {N{H_3}} \right) &= c \cdot V\\n\left( {N{H_3}} \right) &= 1.00\frac{{mol}}{L} \cdot 0.025L\\n\left( {N{H_3}} \right) &= 0.025mol\\pOH &= p{K_b} + \log \left( {\frac{{n\left( {NH_4^ + } \right)}}{{n\left( {N{H_3}} \right)}}} \right)\end{align}\)

\(pOH = - log\left( {1.8 \times 1{0^{ - 5}}} \right) + log\left( {\frac{{0.10\;mol}}{{0.025\;mol}}} \right)\)

\(\begin{align}pOH &= 5.35\\pH &= 14 - pOH\\pH &= 14 - 5.35\\pH &= 8.65\end{align}\)

The pH of the buffer solution is 8.65.

b) The solution is basic because its pH is above 7(8.65).

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 \(N{H_3}\)molecules and form \(NH_4^ + \)ions. The amount of \({H^ + }\)ions added is:

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

Now that we know the concentration of the salt, we can calculate its amount using the formula:

\(\begin{align}n &= c \times V\\n &= 0.100\frac{{mol}}{L} \times 0.750\;L\\n &= 0.075\;mol\end{align}\)

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:

\(pOH = p{K_b} + log\left( {\frac{{n\left( {NH_4^ + } \right)}}{{n\left( {N{H_3}} \right)}}} \right)\)

Since\({H^ + }\)ions react with\(N{H_3}\)molecules, their amount drops by\(1.02 \times 1{0^{ - 4}}\)mol while the amount of\(NH_4^ + \)ions rises by\(1.02 \times 1{0^{ - 4}}\)mol after the addition of HCl.

\(\begin{align}pOH &= - log\left( {1.8 \times 1{0^{ - 5}}} \right) + log\left( {\frac{{0.10 + 1.02 \times 1{0^{ - 4}}}}{{0.025 - 1.02 \times 1{0^{ - 4}}}}} \right)\\pOH &= 5.34\\pH &= 14 - pOH\\pH &= 14 - 5.34\\pH &= 8.64\end{align}\)

The pH changes to 8.64 after the addition of \(HCl\)

a) pH=8.65

b) The solution is basic.

c) pH=8.64