Ionic EquilibriumHard

Question

Calculate the degree of hydrolysis of 0.005 M-K2CrO4. For H2CrO4, Ka1 = infinite, Ka2 =$5 \times 10^{- 7}$.

Options

A.0.002
B.0.02
C.0.2
D.0.005

Solution

$\underset{0.005 - x}{CrO_{4}^{2 -}} + H_{2}O \rightleftharpoons \underset{x}{HCrO_{4}^{-}} + \underset{x}{OH^{-}}$

$K_{h} = \frac{K_{w}}{K_{a_{2}}} = 2 \times 10^{- 8}$

$2 \times 10^{- 8} = \frac{x.x}{0.005 - x} \approx \frac{x^{2}}{0.005} \Rightarrow x = 10^{- 5} $$$\therefore h = \frac{10^{- 5}}{0.005} = 0.002$$

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