Ionic EquilibriumHard

Question

Three solutions of strong electrolytes, 25 ml of 0.1 M-HX, 25 ml of 0.1 M-H2Y and 50 ml of 0.1 N-Z(OH)2 are mixed, the pOH of the resulting solution is

Options

A.1.6
B.7.0
C.12.4
D.11.6

Solution

neq of H+ =$\frac{25 \times 0.1}{1000} \times 1 + \frac{25 \times 0.1}{1000} \times 2 = 7.5 \times 10^{- 3}$

neq of OH– =$\frac{50 \times 0.1}{1000} = 5 \times 10^{- 3}$

∴ neq of H+ left = 2.5 × 10–3

$\Rightarrow \left\lbrack H^{+} \right\rbrack = \frac{2.5 \times 10^{- 3}}{100} \times 1000 = 0.025\text{ M}$

∴ PH = – log (0.025) = 1.6 ⇒ POH = 14 – 1.6 = 12.4

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