Ionic EquilibriumHard

Question

A solution contains 4.25 g ammonia per 250.0 ml of solution. Electrical conductivity measurement at 25°C shows that 0.40% of the ammonia has reacted with water. The pH of the solution is (log 2 = 0.3)

Options

A.11.6
B.2.4
C.12.6
D.10.6

Solution

$\left\lbrack OH^{-} \right\rbrack = \frac{0.4}{100} \times \frac{4.25/17}{250/1000} = 0.004\text{ M}$

POH = – log(0.004) = 2.4

∴ PH = 14 – 2.4 = 11.6

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