Tentukan pH dari larutan penyangga asam yang dibuat dari 400 mL CH3COOH 0,1 M dan 100 mL Ba(CH3COO)2 0,1 M.
Ka CH3COOH = 1 × 10−5
Jumlah zat
(1) Mol CH3COOH = 400 mL × 0,1 M = 40 mmol
(2) Mol Ba(CH3COO)2 = 100 mL × 0,1 M = 10 mmol
Reaksi
\begin{equation*}
\begin{array}
& \ce{CH3COOH (aq) & <=> & CH3COO- (aq) & + & H+ (aq)} & \quad {\color {red} \dotso \: (1)} \\\\
{\color{blue} 40 \text{ mmol}} && &&
\end{array}
\end{equation*}
\begin{equation*}
\begin{array}
& \ce{Ba(CH3COO)2 (aq) & -> & Ba^{2+} (aq) & + & CH3COO- (aq)} & \quad {\color {red} \dotso \: (2)}\\\\
{\color{blue} 10 \text{ mmol}} &&&& {\color{blue} 20 \text{ mmol}}
\end{array}
\end{equation*}
Menentukan pH
\begin{equation*} \begin{split} [\ce{H+}] & = \text{Ka} \:.\: \frac {\ce{mol CH3COOH}}{\ce{mol CH3COO-}} \\\\ [\ce{H+}] & = 10^{-5} \:.\: \frac {40}{20} \\\\ [\ce{H+}] & = 2 \times 10^{-5} \end{split} \end{equation*}
\begin{equation*} \begin{split} \ce{pH & = - \log [H+]} \\\\ \ce{pH & = - \log \left[2 \times 10^{-5}\right]} \\\\ \ce{pH & = 5 - \log 2} \end{split} \end{equation*}