(b)
pH before mixing = 7 + 1/2(4.7447 + log0.143)= 8.95
After mixing :
Millimole of acetic acid = 30.1mmol
Millimole of acetate ion = 91.52 - 30.1= 61.42
pH after mixing = 4.7447 + log(61.42/30.1) = 5.05 (Answer)
pH change = 5.05 - 8.95 = -3.896 = -3.90 (Answer)
(d)
millimole of HCl added = 10.0ml×3.01M = 30.1mmol
Millimole of acetate ion = 0.143M×640ml = 91.52mmol
Millimole of acetic acid = 0.143M×640ml = 91.52mmol
pH = pKa + log([acetate]/[acetic acid]) = - log(1.8×10-5) +log(91.52/91.52) = 4.7447
pH before mixing = 4.74 (Answer)
After mixing
Millimole of acetate = 91.52 - 30.1 = 61.42mmol
Millimole of acetic acid = 91.52 + 30.1 = 121.62mmol
pH = 4.7447 + log(61.42/121.62) = 4.448
pH after mixing = 4.45 (Answer)
pH change = 4.448 - 4.7447 = - 0.2967 = - 0.297
= - 0.30 (Answer)
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