Question 1

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If a 30.0 mL of a 0.20 mol/L HOCN is titrated with 0.250 mol/L KOH,

What is the pH of the solution if 10.0 mL KOH is added?

Question 2

Answer:

$1 \: l \: contains \: 0.20 \: moles \: of \: HOCN \\ 0.03 \: l \: will \: contain \: (0.03 * 0.20) \: moles \\ = 0.006 \: moles \: of \: HOCN \\ 1 \: l \: contains \: 0.250 \: moles \: of \: potassium \: hydroxide \\ 0.01 \: l \: will \: contain \: (0.01 * 0.250) \\ = 0.0025 \: moles \: of \: potassium \: hydroxide \\ KOH + HOCN → KCN + H _(2)O \\ k _(b) = \frac{[K {}^( + ) ][OH {}^( - ) ]}{[KOH]} \\ moles \: of \: OH {}^( - ) \: remaining = 0.006 - 0.0025 \\ = 0.0035 \: moles \\ total \: volume = 0.03 + 0.01 \\ = 0.04 \: litres \\ 0.04 \: l \: contains \: 0.0035 \: moles \\ 1 \: l \: will \: contain \: ( (1 * 0.0035)/(0.04) ) \\ = 0.0875 \: mol \: per \: litre \\ [H {}^( + ) ] = \frac{kw}{[OH {}^( - ) ]} \\ = \frac{1 * {10}^( - 14) }{0.0875} \\ = 1.143 * {10}^( - 13) \\ pH = - log(1.143 * {10}^( - 13) ) \\ pH = 12.94$

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