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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.

Answer» <html><body><p></p>Solution :Calculation of dissociation constant of the acid <br/> `CH_(3)CO OH hArr CH_(3)CO O^(-) + H^(+)` <br/> As `pH = 3, :. [H^(+)] = 10^(-3)M, [CH_(3)CO O^(-)]=[H^(+)]=10^(-3)M` <br/> `K_(a) = ([CH_(3)CO O^(-)][H^(+)])/([CH_(3)CO OH])=(10^(-3)xx10^(-3))/(0.1) = 10^(-5)` <br/> (i) When 1/4th of the acid has been neutralized <br/> `{:(,CH_(3)CO OH,+,NaOH,rarr,CH_(3)CO ONa ,+,H_(2)O),("Initial <a href="https://interviewquestions.tuteehub.com/tag/conc-927968" style="font-weight:bold;" target="_blank" title="Click to know more about CONC">CONC</a>.",0.1 M,,,,,,),("After 1/4th <a href="https://interviewquestions.tuteehub.com/tag/neutralization-15018" style="font-weight:bold;" target="_blank" title="Click to know more about NEUTRALIZATION">NEUTRALIZATION</a>",0.1xx(3)/(<a href="https://interviewquestions.tuteehub.com/tag/4-311707" style="font-weight:bold;" target="_blank" title="Click to know more about 4">4</a>),,,,0.1xx(1)/(4),,),(,=0.075 M ,,,,=0.025 M ,,):}` <br/> `:. pH = pK_(a) + <a href="https://interviewquestions.tuteehub.com/tag/log-543719" style="font-weight:bold;" target="_blank" title="Click to know more about LOG">LOG</a>. (["Salt"])/(["Acid"])=-log 10^(-5) + log. (0.025)/(0.075) = 5 - 0.4771 = 4.5229` <br/> (<a href="https://interviewquestions.tuteehub.com/tag/ii-1036832" style="font-weight:bold;" target="_blank" title="Click to know more about II">II</a>) When 3/4th of the acid has been neutralized <br/> `{:(,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,,):}` <br/> `:. pH = - log 10^(-5) + log. (0.075)/(0.025) = 5 + 0.4771=5.4771`</body></html>


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