Question

Calculate pH for a weak acid/strong base titration. Determine the pH during the titration of 67.3 mL of 0.419 M hypochlorous acid (K-3.5x10-) by 0.419 M NaOH at the following points. (a) Before the addition of any NaOH (b) After the addition of 15.0 mL of NaOH (c) At the half-equivalence point (the titration midpoint) (d) At the equivalence point (e) After the addition of 101 mL of NaOH

Answer 1

(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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