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Question

Calculate the pH of a solution of 0.10 M acetic acid after 50.0 mL of 0.10 M acetic acid solution is treated with 25.0 mL of 0.10 M NaOH. (Ka of acetic acid = 1.8×105)


A

5.6

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B

4.74

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C

8.62

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D

12.8

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Solution

The correct option is B

4.74


CH3COOH+H2OH3O+CH3COOKa=[H3O+][CH3COO][CH3COOH]=1.8×105Before treatment[H3O+]=[CH3COO]=x[CH3COOH]=0.10Ka=[H3O+][CH3COO][CH3COOH]=1.8×1051.8×105=x×x0.10x20.10=1.8×105x2=1.8×106x=[H3O+]=1.3×103pH=2.87

After treatment : Half the acid is neutralized and half remains both solute are now in 75.0ml of solution.

=[CH3COO]=0.050×5075y=0.033y.[CH3COOH]=0.050×5075y=0.033Ka=[H3O+][CH3COO][CH3COOH]=1.8×105=1.8×105=[H3O+]×0.033y0.033y

[H3O+]=1.8×105 M

thus pH = 4.74


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