A 52.0-ml volume of 0.35 m ch3cooh (ka=1.8*10-5) is titrated with 0.40 m naoh. calculate the ph after the addition of 17.0 ml of naoh. express your answer numerically.

Question

Alejandro Randall

Chemistry

3 July, 20:19

A 52.0-ml volume of 0.35 m ch3cooh (ka=1.8*10-5) is titrated with 0.40 m naoh. calculate the ph after the addition of 17.0 ml of naoh. express your answer numerically.

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Answers (1)

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Yadiel Francis · Answer 1

2.76.

Explanation:

Since, the no. of millimoles of CH₃COOH is more than that of NaOH, the medium will be acidic.

C of acid = [ (MV) CH₃COOH - (MV) NaOH] / Vtotal.

C of acid = [ (0.35 M) (52.0 mL) - (0.4 M) (17.0 mL) ] / (69.0 mL) = 0.165 M.

∵ [H⁺] = √ (Ka. C)

∴ [H⁺] = √ (1.8 x 10⁻⁵) (0.165 M) = 1.72 x 10⁻³.

∵ pH = - log[H⁺] = - log (1.72 x 10⁻³) = 2.76.