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Question

20 mL of 0.2 M sodium hydroxide is added to 50 mL of 0.2 M acetic acid to give 70 mL of the solution. Calculate the pH of the solution.
The dissociation constant for CH3COOH is 1.8×105.


A
3.56
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B
4.56
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C
5.56
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D
6.56
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Solution

The correct option is B 4.56
We know,
millimoles = molarity × volume (mL)
millimoles of NaOH=0.2×20=4
millimoles of CH3COOH=0.2×50=10

                     CH3COOH(aq)+NaOH(aq)CH3COONa(aq)+H2O(l)Initial:            10                             4                            0                        0Equilibrium:(104)                    0                             4                         4

[CH3COONa]=470 M[CH3COOH]=670 M
pKa=log(Ka)pKa=log(1.8×105)pKa=(50.26)=4.74
pH=pKa+log[salt][acid]pH=pKa+log([CH3COONa][CH3COOH])pH=4.74+log470670pH=4.74+log(0.67)pH=4.740.18=4.56pH=4.56

Chemistry

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