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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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12.176

1.842

13.301

0.699

answer is D.

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Detailed Solution

Following reaction take place

H₂SO_{4 +}

KOH

\to

K₂SO₄ + H₂O

Milli

equivalents

50 x 0.2 = 10

100 x 0.4 = 40

0 0

After reaction

10 - 10

40 - 10 = 30

10 10

${[{OH}^{-}]}_{Final} = \frac{30 \times 10^{- 3}}{2} = 15 \times 10^{- 3} = 1.5 \times 10^{- 2}$

P^OH = 2 - log(1.5) = 2 - 0.176 = 1.824

P^H = 14 - 1.824 = 12.176

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