Question

Consider the titration of 0.050 M HCl with 0.050 M NaOH. Calculate the pH: (a) of the pure HCl solution; (b) of 25.00 mL acid plus 12.50 mL base; (c) at the quivalence point; and (d) of 25.00 mL acid and 37.50 mL base.

Answer 1

a)   [HCl] = 0.05 M

    pH = -log [H+] = - log (0.05) = 1.3

Therefore,

pH of pure HCl = 1.3

b) 25.00 mL acid plus 12.50 mL base

   [HCl] = 0.05 M x 0.025 L = 0.00125 mol

   [NaOH] = 0.05 M x 0.0125 L = 0.000625 mol

Hence, [HCl] > [NaOH]

Then,

pH = - log {[HCl] - [NaOH] / total volume in Litres}

      = - log {0.00125 mol - 0.000625 mol/ 0.025 L + 0.0125 L }

      = 1.78

Therefore,

pH = 1.78

c) For strong acid vs strong base base titrations,

     pH at equivalence point = 7

d) 25.00 mL acid and 37.50 mL base

   [HCl] = 0.05 M x 0.025 L = 0.00125 mol

   [NaOH] = 0.05 M x 0.0375 L = 0.001875 mol

Hence, [NaOH] > [HCl]

Then,

pOH = - log { [NaOH] - [HCl] / total volume in Litres}

      = - log { 0.001875 mol - 0.00125 mol / 0.025 L + 0.0375 L }

      = 2

Hence,

pH = 14 - pOH = 14- 2 = 12

Therefore,

pH = 12