(a) before the addition of any base we have only weak acid whose pH is given by
pH = 1/2 ( pKa - log [acid) )
pKa = -log Ka = -log ( 3.5*10^-8) = 7.45
pH = 1/2 ( 7.45 - log (0.419) ) = 3.91
(b) After addition of 15 mL of NaOH
Net Reaction is
Acid + Base = Salt + H2O
initial moles of acid = molarity * volume = 0.419 * 67.3 = 28.2 mmol
moles of NaOH added = 6.285 mmol
hence mmols of acid remains = 28.2 -6.285 = 21.915 mmol
mmols of salt formed = 6.285
hence we are left with weak acid and its salt which is nothing but buffer whose pH is given by
pH = pKa + log [Salt] / [Acid]
pH = 7.45 + log [6.285] / [21.915]
pH = 6.9
(c) At Midpoint concentration of acid and its strong conjugate basic salt is same hence from the above equation
when [acid] = [salt]
then pH = pKa
therefore pH = 7.45
(d) At equivalence point
mmols of acid and base are equal
hence 28.2 mmols of acid react with 28.2 mmol of base to form 28.2 mmol of salt
hence we are left with salt of weak acid and strong base whose pH is given by
pH = 1/2 ( pKw + pKa + log [salt])
[salt] = moles / volume = 28.2 mmols / (67.3 + 67.3) = 0.209 M
pH = 1/2 ( 14 + 7.45 + log (0.209) ) = 10.38
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