Question

When a 19.8 mL sample of a 0.460 M aqueous acetic acid solution is titrated with a 0.398 M aqueous sodium hydroxide solution, what is the pH at the midpoint in the titration? pH =

Answer 1

(a) Ka for acetic acid = 1.8 x 10-5 pKa=-log(1.8 x 10-5) = 4.74 At mid point, the number of moles of acid (CH3COOH) is equal to its conjugate base (CH3C00-). So, pH = pka = 4.74 pH of the solution at mid point = 4.74

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(b) Before any addition of base : Conc. HE + H20 F 1 F + H30* Initial 0.430 0 Change - -x + Equilibrium * x.x wa -= 3.5x104 0.430 - x x² +3.5x10+x-1.51x10+ = 0 by solving we get, x=0.012094 [H*]=0.012094M pH =-log[H* ]=- log(0.012094)=1.92 pH=1.92