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 = 1.8×10−5)

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+H2O→H3O+CH3COO−Ka=[H3O+][CH3COO−][CH3COOH]=1.8×10−5Before treatment[H3O+]=[CH3COO−]=x[CH3COOH]=0.10Ka=[H3O+][CH3COO−][CH3COOH]=1.8×10−51.8×10−5=x×x0.10x20.10=1.8×10−5x2=1.8×10−6x=[H3O+]=1.3×10−3pH=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]=1.8×10−5=1.8×10−5=[H3O+]×0.033y0.033y [H3O+]=1.8×10−5 M thus pH = 4.74

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