Question

Consider what is the pH of the solution when 25.0mL of 0.01M NaOH is added to 25.0mL of 0.01M acetic acid. Is it the same as in the case of HCl solution? What will the pH of the solution be if exactly 12.50mL of 0.01M NaOH solution is added to 25.0mL of 0.01M acetic acid solution?

Answer 1

CH3COOH + OH- (aq) <----> CH3COO- (aq) + H2O ( l)

acetic acid moles = Mx V = 0.01 x 25/1000 = 0.00025

NaOH moles = OH- moles = 0.01 x 25/1000 = 0.00025

hence all acetic acid is converted to acetate , now acetate combines with water to form acteic acid and Oh- to some extent

CH3COO-(aq) + H2O (l) <----> CH3COOH + OH- , Kb = Kw/Ka of Ch3COOH = 5.56 x 10^ -10

ar equilibrium CH3COO- = 0.0005-X , CH3cOOH = OH- moles = X , vol of solution = 50 ml = 0.05 L

Kb = [CH3COOH] [OH-]/[CH3COO-]

5.56 x 10^ -10 = ( X/0.05) ( X/0.05) / ( 0.00025-X) /0.05

X = 8.33 x 10^ -8 ,   [OH-] = 8.33x10^ -8 / 0.05 = 1.67 x 10^ -6 M

pOH = -log [OH-]= -log ( 1.67x10^ -6) = 5.78 , pH = 14-pOH = 14-5.78 = 8.22

2) HCl is strong acid and dissociatesly completly to give H+ ions , when equal moles of strong acid and strong base added we get neutral solution , hence pH = 7

3) OH- moles = M x V = 0.01 x 12.5/1000 = 0.000125

after reaction with OH- , acetic acid moles = 0.00025-0.000125 = 0.000125

moles of acetate formed = moles of OH- reacted = 0.000125

vol = 25+12.5 = 37.5 ml = 0.0375 L

now we had weak acid and its conjugate base acetate , hence we had buffer

for buffer pH = pka + log [conjugate base] /[acid]    wher pka of acid = 4.745

pH = 4.745 + log ( 0.000125/0.0375) / ( 0.000125/0.0375)

       = 4.745