1.

Calculate the pH of a solution obtained by mixing 100 ml of 0.10 M CH3COOH and 100 ml of 0.05 M NaOH solution. Ka for CH3COOH is 1∙8 x 10-5 ?

Answer»

NaOH + CH3COOH → CH3COONa + H2O

100 ml of 0 ∙ 10 M CH3COOH contains

= 100 × 0∙1

= 10 m mol CH3COOH

100 ml of 0∙5 M NaOH contains

= 100 × 0∙05

= 5 m mol of NaOH

5 m mol NaOH react with 5 m mol CH3COOH to form 5 mol of CH3COONa.

5 m mol of CH3COOH remain unreacted. After mixing,

[CH3COOH] = \(\frac{5}{200}\) M = 2.5 x 10-2 M

[CH3COOH] = \(\frac{5}{200}\) = 2.5 x 10-2 M

pH = pKa + log\(\frac{[CH_3COO^-]}{[CH_3COOH]}\)

pH = pKa + log\(\frac{[2.5\times10^{-2}]}{[2.5\times10^{-2}]}\)

Ka = 1∙8 × 10−5

pH = pKa

= − log (1 ∙ 8 × 10−5)

= 4∙74.


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