Question

A 50.0-mL solution of 0.100 M potassium hydrogen phthalate (KHP; (HP– has Ka ~ 4.0 x 10–6 ) is titrated with 0.100 M of NaOH. Calculate the pH of the solution at various points of the titration: (a) Before 0.100 M NaOH is added; (b) After 25.0 mL of NaOH is added; (c) After 50.0 mL of NaOH is added, and (d) After 75.0 mL of NaOH is added

Answer 1

a) HP- + H2O --------> P2- + H3O+

Ka = [ P2- ] [ H3O+]/[HP- ] = 3.98×10^-6

[P2- ][H3O+]/[HP-] = 3.98×10^-6

at equillibrium

[ P2- ] = X

[ H3O+ ] = X

[ HP- ] = 0.1-X

Therefore,

X^2/(0.1-X) = 3.98×10^-6

x value is very small ,So we can assume 0.1 - X as 0.1

X^2/0.1 = 3.98×10^-6

X^2 = 3.98 × 10^-7

X = 6.31×10^-4

Therefore,

[ H3O+ ] = 6.31×10^-4M

pH= - log [ H3O+ ]

= - log ( 6.31×10^-4)

= 3.20

b) HP- + OH- -------> P2- + H2O

stoichiometrically, 1mole of OH- react with 1mole HP- to form 1mole of P2-

25 ml is Half equivalence point

at Half neutalization point , pH = pKa

pKa of KHP = 5.4

Therefore,

pH = 5.4

c) at equivalence point

P2- + H2O <------> HP- + OH-

Kb = [ OH- ] [ HP- ]/[P2- ]

Kb = Kw/Ka

= 1× 10^-14/3.98×10^-6

= 2.51 × 10^-9

Concentration of P2- = 0.1M/2 = 0.05M

Therefore,

X^2/(0.05-X ) = 2.51×10^-9

X^2/0.05 = 2.51 × 10^-9

X^2 = 1.26×10^-10

X = 1.12 × 10^-5

[ OH- ] = 1.12×10^-5

pOH = - log [ OH- ]

= - log (1.12×10^-5)

= 4.95

pH = 14 - pOH

= 14 - 4.95

= 9.05

d) 25ml of excess NaOH added

moles of OH- by NaOH =( 0.1mol/1000ml)×25ml = 0.0025mole

Total volume = 125ml

[ OH- ] =(0.0025mol/125ml)×1000ml = 0.02M

pOH =1.70

pH = 14 - 1.70

= 12.3