Question

A 37.00 mL solution of 0.390 M weak acid Ha(aq) (Ka=2.25x10^-5) is titrate with 0.390 M NaOH(aq). Calculate the pH of the solution 14.00 mL past the end point.

Answer 1

According law of milliequivalence

N1V1 = N2 V2

N1 = normality of acid = 0.39N (SInce Normaility = Molarity for monobasic acid)

V1 = Volume of acid consumed for 14ml of 0.39N NaOH = ?

N2 = Normality of NaOH = 0.39N (since normality = molarity for mono acidic base)

V2 = Volume NaOH added = 14 ml

V1 = N2 V2 / N1 = 0.39 * 14 / 0.39 = 14 ml

So volume of Acid left is 37 - 14 = 23ml

23ml of 0.39M acid is diluted to a volume 37+14ml = 51ml

From law of dilution C1V1 = C2V2

C1 = Concentration of acid = 0.39M

V1 = VOlume left after reacting with NaOH = 23ml

C2= Concentration of acid after reacting with NaOH = ?

V2 = VOlume of solution = 51ml

C2 =0.39*23 / 51 = 0.17588M

Since it is a weak acid its H⁺ concentraation depends on degree of dissociation

                                          HA                                        H⁺ +                 A^-

Initial                                C                    0          0

At equillibrium                C(1-x)                                       Cx      Cx

Ka = C²x² / C(1-x)     as x is very small for weak acids 1-x can be neglected

Cx = Ka

      = 2.25x10^-5

        = 4.743 x 10^-3

So H⁺ concentration is 4.743x10^-3 M

pH = - log [H⁺] = - log 4.743x10^-3 = -log 4.743 + 3 = 3 - 0.6761 = 2.3239