Ionic EquilibriumMedium
Question
An amount of 0.1 mole of CH3NH2 (Kb = 5 × 10–4) is mixed with 0.08 mole of HCl and diluted to one litre. What will be the H+ concentration in the solution?
Options
A.1.25 × 10–4 M
B.8 × 10–11 M
C.1.6 × 10–11 M
D.2 × 10–3 M
Solution
CH3NH2 + H+ $\rightleftharpoons$ CH3NH3+
0.1 mole 0.08 mole 0
Final 0.02 mole 0 0.08 mole
$P^{OH} = P^{K_{b} + \log\frac{\left\lbrack CH_{3}NH_{3}^{+} \right\rbrack_{0}}{\left\lbrack CH_{3}NH_{2} \right\rbrack_{0}}} = - \log\left( 5 \times 10^{- 4} \right) + \log\frac{0.08}{0.02} $$$\therefore\left\lbrack OH^{-} \right\rbrack = \frac{5 \times 10^{- 4}}{4} \Rightarrow \left\lbrack H^{+} \right\rbrack = \frac{10^{- 14}}{\left\lbrack OH^{-} \right\rbrack} = 8 \times 10^{- 11}\text{ M}$$
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