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Question

Calculate the pH of a solution obtained by mixing 2 mL of HCl of pH 2 and 3 mL of solution of KOH of pH = 12


A

10.30

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B

3.70

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C

11.30

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D

None of these

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Solution

The correct option is C

11.30


H+ in HCl solution (pH=2) = 102M; [OH] in KOH solution (pOH = 14 - 12 = 2) = 102M

Excess m mol of OH in 5 mL mixture = (3 × 102) (2 × 102)=(1.0 × 102)

[OH] in mixture = (1.0 × 102)5=2.0 × 103M;

pOH = log2 ×103=3 log2;

pH=14 (3log2)=11.30


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