Question 17.SIE.1: A sample of 1.25 L of HCl gas at 21 °C and 96.3 kPa is bubb...
A sample of 1.25 L of HCl gas at 21 °C and 96.3 kPa is bubbled through 0.500 L of 0.150 M NH_3 solution. Calculate the pH of the resulting solution assuming that all the HCl dissolves and that the volume of the solution remains 0.500 L.
The number of moles of HCl gas is calculated from the ideal gas law:
n=\frac{P V}{R T}=\frac{(96.3 \,kPa )(1.25 \,\cancel{L} )}{(8.314\, L – kPa / mol – \cancel{K} )(294 \,\cancel{K} )}=0.0492 \,mol \,HCl
The number of moles of NH_3 in the solution is given by the product of the volume of the solution and its concentration:
\text{Moles}\,NH _3=(0.500 \,\cancel{L} )\left(0.150 \,mol \,NH _3 / \cancel{L} \right)=0.0750 \,mol \,NH _3
The acid HCl and base NH_3 react, transferring a proton from HCl to NH_3, producing NH_4^+ and Cl^- ions:
HCl (g)+ NH _3(a q) \longrightarrow NH _4^{+}(a q)+ Cl ^{-}(a q)
To determine the pH of the solution, we first calculate the amount of each reactant and each product present at the completion of the reaction. Because you can assume this neutralization reaction proceeds as far toward the product side as possible, this is a limiting reactant problem.
\begin{array}{|l|c|c|c|c|}\hline& HCl (g)+& NH _3(a q)\longrightarrow & NH _4^{+}( aq ) & + Cl ^{-}(a q) \\\hline \begin{array}{l}\text{Before reaction}\\\text{(mol)}\end{array}& 0.0492 & 0.0750 & 0 & 0 \\\hline \begin{array}{l}\text{Change (limiting}\\\text{reactant) (mol)}\end{array}& -0.0492 & -0.0492 & +0.0492 & +0.0492 \\\hline \begin{array}{l}\text{After reaction}\\\text{(mol)}\end{array}& 0 & 0.0258 & 0.0492 & 0.0492 \\\hline\end{array}
Thus, the reaction produces a solution containing a mixture of NH_3, NH_4^+, and Cl^-. The NH_3 is a weak base \left(K_b=1.8 \times 10^{-5}\right), NH_4^+ is its conjugate acid, and Cl^- is neither acidic nor basic. Consequently, the pH depends on [NH_3] and [NH_4^+]:
\begin{aligned}&{\left[ NH _3\right]=\frac{0.0258 \,mol\, NH _3}{0.500\, L \,\text{solution}}=0.0516 \,M}\\&{\left[ NH _4^{+}\right]=\frac{0.0492\, mol\, NH _4^{+}}{0.500 \,L \,\text{solution}}=0.0984 \,M}\end{aligned}
We can calculate the pH using either K_b for NH_3 or K_a for NH _4^{+} . Using the K_b expression, we have:
\begin{array}{|l|c|c|c|c|}\hline & NH _3(a q)+& H _2 O (l) \rightleftharpoons & NH _4^{+}(a q) & + OH ^{-}(a q) \\\hline \text{Initial}(M) & 0.0516 & – & 0.0984 & 0 \\\hline \text{Change}(M) & -x & – & +x & +x \\\hline \text{Equilibrium}(M) & (0.0516-x) & – & (0.0984+x) & x \\\hline\end{array}
\begin{aligned}K_b &=\frac{\left[ NH _4^{+}\right]\left[ OH ^{-}\right]}{\left[ NH _3\right]}=\frac{(0.0984+x)(x)}{(0.0516-x)}\cong \frac{(0.0984) x}{0.0516}=1.8 \times 10^{-5}\\x &=\left[ OH ^{-}\right]=\frac{(0.0516)\left(1.8 \times 10^{-5}\right)}{0.0984}=9.4 \times 10^{-6}M\end{aligned}
Hence, pOH=-\log \left(9.4 \times 10^{-6}\right)=5.03
and pH = 14.00 – pOH = 14.00 – 5.03 = 8.97.