Question

a) . Calculate the pH of a titration of 100 mL 1.5M HCOOH with 1.25M NaOH at equivalence. WHY is the pH what it is (if it’s 7, why is the solution neutral? If not 7, why?)

b) . If it takes 50 mL of 2.5M NaOH to titrate 60 mL HF to equivalence, what is the concentration of the HF?

c) What is true at the half-equivalence point of a titration of acid with base in terms of moles of base added relative to moles of acid? In terms of the concentrations of the acid and its conjugate base? In terms of pH of solution?

Answer 1

a) pH of acidic buffer = pka + log(HcooNa/HCOOH)

pka of HCOOH = -logKa = -log(1.8*10^-4) = 3.74

no of mol of HCOOH = 100*1.5/1000 = 0.15 mol

No of mol of HCOONa = NaOH added = 0.15 mol

volume of NaOH must be added to reach euivalence point = n/M*1000 = 0.15/1.25*1000

                                                       = 120 ml
HCOOH + NaOH ---> HCOONa + H2O

HCOONa = salt of weak acid,strong base = it is basic

Concentration of HCOONa = n/v = 0.15/(0.22) = 0.68 M

pH = 7+1/2(pka+logC)

   = 7+1/2(3.74+log0.68)

   = 8.8

b) no of mol of NaOH = M*V

                     = 2.5*(50/1000) = 0.125 mol

To reach equivalence point

no of mol of HF required = 0.125 mol

concentration of HF = n/v = 0.125/60*1000 = 2.08 M

c) at the half-equivalence point of a titration of acid with base

No of mol of base added will be half of the No of mol of acid taken.

The concentration of acid and its conjugate base will be same.

The pH of solution = pka of weak acid.