Question

A mixture contains benzoic acid, a monoprotic acid (Ka = 6.28 x 10-5, 0.250 M), and sodium benzoate (0.400 M). 500 mL of this buffer were treated with 250 mL of a NaOH solution whose pH was 12.875. What should the pH be after addition of the base?

Answer 1

Given that, concentration of benzoicacid = 0.250M concentration of sodium benzoate = 0.400 M volume of buffer = 500 mL volume of NaOH = 250 mL pH of NaOH solution = 12.875 of monoprotic acid (benzoic acid) = 6.28 x 10 PK, of benzoicacid = 4.20 K Calculate the pH of buffer using Henderson-Hasselbalch equation as follows: Benzoate pH = PK, +log [Benzoic acid] 0.400 = 4.20+log $ 0.250 = 4.40 Calculate concentration of OH as follows: pH = 14-POH =14-12.875 =1.125 pOH = -log[OH-] TOH"]=0.075 M Calculate millimole of NaOH as follows: Number of millimoles of NaOH = concentration x volume =0.075 MX 250 mL =18.75 millimoles Calculate millimoles of buffer solution as follows:

Millimoles of benzoic acid = concentration x volume =0.25M X 500 mL =125 millimoles Millimoles of bezoate = concentration x volume =0.4 MX 500 mL =200 millimoles Calculate the pH of buffer after addition of 18.75 millimoles of NaOH using Henderson-Hasselbalch equation as follows: pH of buffer=pK Benzoate + NaOH +log [Benzoic acid – NaOH] 200+18.75 = 4.20+ log (125-18.75] = 4.51 The change in pH of buffer after addition of NaOH is, Change in buffer = 4.51-4.40 = 0.11 Therefore, the change in pH after addition of base is 0.111.