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0.1 M CH_(3)CO OH (pH = 3) is titrated with 0.05 M NaOH solution. Calculate the pH when (i) 1/4th of the acid has been neutralized.(ii) 3/4th of the acid has been neutralized. |
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Answer» Solution :Calculation of dissociation constant of the acid `CH_(3)CO OH hArr CH_(3)CO O^(-) + H^(+)` As `pH = 3, :. [H^(+)] = 10^(-3)M, [CH_(3)CO O^(-)]=[H^(+)]=10^(-3)M` `K_(a) = ([CH_(3)CO O^(-)][H^(+)])/([CH_(3)CO OH])=(10^(-3)xx10^(-3))/(0.1) = 10^(-5)` (i) When 1/4th of the acid has been neutralized `{:(,CH_(3)CO OH,+,NaOH,rarr,CH_(3)CO ONa ,+,H_(2)O),("Initial CONC.",0.1 M,,,,,,),("After 1/4th NEUTRALIZATION",0.1xx(3)/(4),,,,0.1xx(1)/(4),,),(,=0.075 M ,,,,=0.025 M ,,):}` `:. pH = pK_(a) + LOG. (["Salt"])/(["Acid"])=-log 10^(-5) + log. (0.025)/(0.075) = 5 - 0.4771 = 4.5229` (II) When 3/4th of the acid has been neutralized `{:(,CH_(3)CO OH,+,NaOH,rarr,CH_(3)CO ONa , +,H_(2)O),("Initial conc.",0.1 M,,,,,,),("After 3/4th",0.1xx1/4M,,,,0.1xx3/4M,,),("neutralization",=0.025 M,,,,=0.075 M,,):}` `:. pH = - log 10^(-5) + log. (0.075)/(0.025) = 5 + 0.4771=5.4771` |
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