Answers: 2
![1 \: l \: contains \: 0.20 \: moles \: of \: HOCN \\ 0.03 \: l \: will \: contain \: (0.03 \times 0.20) \: moles \\ = 0.006 \: moles \: of \: HOCN \\ 1 \: l \: contains \: 0.250 \: moles \: of \: potassium \: hydroxide \\ 0.01 \: l \: will \: contain \: (0.01 \times 0.250) \\ = 0.0025 \: moles \: of \: potassium \: hydroxide \\ KOH + HOCN β KCN + H _{2}O \\ k _{b} = \frac{[K {}^{ + } ][OH {}^{ - } ]}{[KOH]} \\ moles \: of \: OH {}^{ - } \: remaining = 0.006 - 0.0025 \\ = 0.0035 \: moles \\ total \: volume = 0.03 + 0.01 \\ = 0.04 \: litres \\ 0.04 \: l \: contains \: 0.0035 \: moles \\ 1 \: l \: will \: contain \: ( \frac{1 \times 0.0035}{0.04} ) \\ = 0.0875 \: mol \: per \: litre \\ [H {}^{ + } ] = \frac{kw}{[OH {}^{ - } ]} \\ = \frac{1 \times {10}^{ - 14} }{0.0875} \\ = 1.143 \times {10}^{ - 13} \\ pH = - log(1.143 \times {10}^{ - 13} ) \\ pH = 12.94](/tpl/images/2384/5667/af5ff.png)
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If a 30.0 mL of a 0.20 mol/L HOCN is titrated with 0.250 mol/L KOH,
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