Question 17.11: Calculating the pH of a Buffer from Given Volumes of Solutio...
Calculating the pH of a Buffer from Given Volumes of Solution
Instructions for making up a buffer say to mix 60 . \mathrm{mL}(0.060 \mathrm{~L}) of 0.100 M \mathrm{NH}_{3} with 40. mL (0.040 L) of 0.100 M NH_4Cl. What is the pH of this buffer?
PROBLEM STRATEGY
The buffer contains a base and its conjugate acid in equilibrium. The equation is
\mathrm{NH}_{3}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons \mathrm{NH}_{4}^{+}(a q)+\mathrm{OH}^{-}(a q)
The problem is to obtain the concentration of \mathrm{H}_{3} \mathrm{O}^{+}or \mathrm{OH}^{-}in this equilibrium mixture. To do the equilibrium calculation, you need the starting concentrations in the solution obtained by mixing the \mathrm{NH}_{3} and \mathrm{NH}_{4} \mathrm{Cl} solutions. For this, you calculate the moles of \mathrm{NH}_{3} and moles of \mathrm{NH}_{4}{ }^{+}added to the buffer solution, then divide by the total volume of buffer. Then you are ready to do the equilibrium calculation.
How many moles of \mathrm{NH}_{3} are added? Recall that
\text { Molarity } \mathrm{NH}_{3}=\frac{\text { moles } \mathrm{NH}_{3}}{\text { liters } \mathrm{NH}_{3} \text { solution }}
Note that the instructions say to add 60 . \mathrm{mL} (or 0.060 \mathrm{~L} ) of 0.100 \mathrm{M} \mathrm{NH}_{3}. So
\begin{aligned} \text { Moles }\,\mathrm{NH}_{3} & =\text { molarity } \mathrm{NH}_{3} \times \text { liters } \mathrm{NH}_{3} \text { solution } \\ & =0.100 \frac{\mathrm{mol}}{\mathrm{L}} \times 0.060 \mathrm{~L}=0.0060 \mathrm{~mol} \mathrm{NH}_{3} \end{aligned}
In the same way, you find that you have added 0.0040 \mathrm{~mol} \mathrm{NH}_{4}^{+}(from \mathrm{NH}_{4} \mathrm{Cl} ). Assume that the total volume of buffer equals the sum of the volumes of the two solutions.
\text { Total volume buffer }=60 . \mathrm{mL}+40 . \mathrm{mL}=100 . \mathrm{mL}(0.100 \mathrm{~L})
Therefore, the concentrations of base and conjugate acid are
\begin{aligned} {\left[\mathrm{NH}_{3}\right] } & =\frac{0.0060 \mathrm{~mol}}{0.100 \mathrm{~L}}=0.060 \mathrm{M} \\ {\left[\mathrm{NH}_{4}{ }^{+}\right] } & =\frac{0.0040 \mathrm{~mol}}{0.100 \mathrm{~L}}=0.040 \mathrm{M} \end{aligned}
STEP 1 Fill in the concentration table for the acid-base equilibrium (base ionization of \mathrm{NH}_{3} ).
STEP 2 You substitute the equilibrium concentrations into the equilibrium-constant equation.
\begin{aligned} \frac{\left[\mathrm{NH}_{4}{ }^{+}\right]\left[\mathrm{OH}^{-}\right]}{\left[\mathrm{NH}_{3}\right]} & =K_{b} \\ \frac{(0.040+x) x}{0.060-x} & =1.8 \times 10^{-5} \end{aligned}
STEP 3 To solve this equation, you assume that x is small compared with 0.040 and 0.060 . So this equation becomes
\frac{0.040 x}{0.060}=1.8 \times 10^{-5}
Therefore, x=(0.060 / 0.040) \times\left(1.8 \times 10^{-5}\right)=2.7 \times 10^{-5}. (Check that x can be neglected in 0.040+x and 0.060-x.) Thus, the hydroxide-ion concentration is 2.7 \times 10^{-5} M. The \mathrm{pH} of the buffer is
\mathrm{pH}=14.00-\mathrm{pOH}=14.00+\log \left(2.7 \times 10^{-5}\right)=\mathbf{9 . 4 3}
(If you have 2.7 \times 10^{-5} in your calculator from the previous calculation, you obtain the \mathrm{pH} by pressing the \log button, then adding 14.)