Ionic EquilibriumHard
Question
An amount of 0.16 g of N2H4 is dissolved in water and the total volume is made up to 500 ml. What is the percentage of N2H4 that has reacted with water in this solution? The value of Kb for N2H4 = 4.0 × 10−6.
Options
A.0.02%
B.0.014%
C.2%
D.2.82%
Solution
$\left\lbrack W_{2}H_{4} \right\rbrack_{O} = \frac{0.16/32}{500/1000} = 0.01\text{ M}$
$\therefore\alpha = \sqrt{\frac{4 \times 10^{- 6}}{0.01}} = 0.02\text{ or 2\%}$
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