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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

12.176

b

1.842

c

13.301

d

0.699

answer is D.

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Detailed Solution

Following reaction take place

     H2SO4 +KOH→K2SO4 + H2O

Milli

equivalents

50 x 0.2 = 10100 x 0.4 = 40         0             0
After reaction10 - 1040 - 10 = 30       10             10

OH-Final =30×10-32=15×10-3=1.5×10-2

POH = 2 - log(1.5) = 2 - 0.176 = 1.824

PH = 14 - 1.824 = 12.176

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