Question

3. A 60.0 mL solution contains 0.300 M of a triprotic acid (H3A, pK1 = 3.05, pK2 = 7.89, pK2 = 11.06). What volume of titrant (0.1546 M sodium hydroxide) is needed to achieve a solution in which [HA2–] = [A3–]? Report your answer in mL.

Answer 1

Initially all H3A needed to convert to H2A -

H3A + OH- ----> H2A- (aq) + H2O

hence NaOH moles required = moles of H3A = M x V = 0.3 x 60/1000 = 0.018

H2A- (aq) + OH- (aq) ---> HA2- (aq) + H2O ( l)

now all H2A needed to convert to HA2- , moles of NaoH required = 0.018

HA2-(aq) + OH- (aq) ----> A3-(aq) + H2O (l)

now half or HA2- is needed to convert to A3- , hence OH- moles needed = 0.0018/2 = 0.009

now total OH_ moles needed = 0.018+0.018+0.009 = 0.045

M of NaoH = moles/ vol in L

0.1546 = 0.045/vol in L

vol of NaOH = 0.045/0.1546 = 0.291 L = 291 ml