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Q.

50 ml of 0.2N H2SO4 is mixed with 100 ml of 0.4N KOH solution and 1.85 lit of distilled water is added. The pH of resulting solution is (log 1.5 = 0.176)

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a

13.301

b

0.699

c

1.842

d

12.176

answer is D.

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Detailed Solution

Following reaction take placeMilliequivalentsOH-Final =30×10-32=15×10-3=1.5×10-2POH = 2 - log(1.5) = 2 - 0.176 = 1.824PH = 14 - 1.824 = 12.176
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