Question

A buffer solution was prepared by adding 4.95 g of sodium acetate to 250 mL of 0.15 M acetic acid.

What is the pH of 100 mL of this buffer solution if you add 82 mg of NaOH to the solution

Answer 1

pH of acidic buffer = pka + log[C2H3O2-]/[HC2H3O2]

      pka = - logKa

          = -log(1.8*10^-5)

          = 4.74

No of mol of HC2H3O2(aceticacid) = 250*0.15/1000 = 0.0375 mol

No of mol of NaC2H3O2 present = w/mwt = 4.95/82 = 0.0604 mol

in 100 ml buffer,

No of mol of HC2H3O2(aceticacid) = 0.0375*100/250 = 0.015 mol

No of mol of NaC2H3O2 present = 0.0604*100/250 = 0.02416 mol

No of mol of NaOH added =w/mwt = 82*10^-3/40 = 0.00205 mol

pH = pka + log[C2H3O2- + NaOH]/[HC2H3O2-NaOH]

   = 4.74+log((0.02416+0.00205)/(0.015-0.00205))

   = 5.05