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Question: If 50 ml of 0.2 M KOH is added to 40 ml of 0.5 HCOOH, the pH of the resulting solutions is (ka = 1. ...

If 50 ml of 0.2 M KOH is added to 40 ml of 0.5 HCOOH, the pH of the resulting solutions is (ka = 1. 8 x 10-4)

A

3.753.75

B

5.65.6

C

7.57.5

D

3.43.4

Answer

3.753.75

Explanation

Solution

MeqofKOH=50×0.2=10MeqofKOH = 50 \times 0.2 = 10

MeqofHCOOH=40×0.2=20MeqofMeqofHCOOH = 40 \times 0.2 = 20Meqof

MeqofKOHleft=10MeqofHCOOKformed=10∴pH=pKa+log⁡[salt][acid]MeqofKOHleft = 10MeqofHCOOKformed = 10\therefore pH = pK_{a} + \log\frac{\lbrack salt\rbrack}{\lbrack acid\rbrack}

⇒pH=−log⁡ka+log⁡1010⇒pH=−log⁡ka+log⁡1010⇒pH=−log⁡(1.8×10−4)=3.75\Rightarrow pH = - \log k_{a} + \log\frac{10}{10} \Rightarrow pH = - \log k_{a} + \log\frac{10}{10} \Rightarrow pH = - \log\left( 1.8 \times 10^{- 4} \right) = 3.75