First slide

PH of strong acids and strong bases

Question

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)

Moderate

A

13.301
B

0.699
C

1.842
D

12.176

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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