Ionic EquilibriumHard

Question

An aqueous solution initially contains 0.01 M-RNH2 (Kb = 2.0 × 10−6) and 10−4 M-NaOH. The final concentration of OH− in the solution is about

Options

A.10−4 M
B.2.0 × 10−4 M
C.3.0 × 10−4 M
D.1.414 × 10−4 M

Solution

$\underset{0.01 - x}{RNH_{2}} + H_{2}O \rightleftharpoons \underset{x}{RNH_{3}^{+}} + \underset{x + 10^{- 4}}{OH^{-}}$

$2 \times 10^{- 6} = \frac{x\left( x + 10^{- 4} \right)}{0.01 - x} \Rightarrow x = 10^{- 4} $$$\therefore\left\lbrack OH^{-} \right\rbrack = 2 \times 10^{- 4}\text{ M}$$

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