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Question

5.0mL of 0.1MNaOH solution is added to 50mL of the 0.1M acetic acid solution. Calculate the pH of the resulting acetic acid solution.
(Ka=1.8×105)

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Solution

CH3COOH+H2OCH3COO+H3O+
Ka=[CH3COO][H3O+][CH3COOH]
Let [H3O+]=[CH3COO]=x
Ka=x20.1MM2
After adding 5 of NaOH
Moles of CH3COOH=0.050×0.1/=0.005
Moles of NaOH added =0.0050×0.1/=0.0005
CH3COOH+OHCH3COO+H2O
After adding NaOH,0.0005 of CH3COO
and 0.0045 of CH3COOH is left.
Ka=1.8×105=[H3O+][0.0005][0.0045]
[H3O+]=0.000162M
pH=log(0.000162)
=3.79

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