Question

The pK, of hypochlorous acid is 7.530. A 51.0 mL solution of 0.149 M sodium hypochlorite (NaCl) is titrated with 0.301 MHCI. Calculate the pH of the solution after the addition of 10.0 mL of 0.301 M HCI. pH - 6.82 Calculate the pH of the solution after the addition of 26.2 mL of 0.301 M HCI. pH 7.24 Calculate the pH of the solution at the equivalence point with 0.301 M HCI. pH = 7.2246

Answer 1

a)

mmol of OCl- = MV = 51*0.149 = 7.599

mmol of H+ = MV = 0.301*10 = 3.01

then, HOCl forms:

mmol of OCl- left = 7.599-3.01 = 4.589

mmol of HCOl = 0+3.01 = 3.01

pH = pKa + log(A-/HA)

pH = 7.530 + log(4.589/3.01 )

ph = 7.530+0.183 = 7.713

b)

similarly

mmol of OCl- = MV = 51*0.149 = 7.599

mmol of H+ = MV = 0.301*26.2 = 7.8862

excess acid:

mmol of H+ = 7.8862-7.599 = 0.2872

Vtotal = 51+26.2 = 77.2

[H+] = 0.2872/77.2 = 0.00372

pH = -log(0.00372) = 2.429

c)

in equivalence

V = mmol/V = 7.599/0.301 = 25.2458 mL

Vtotal = 25.2458+51 = 76.2458

[HOCl] = 7.599/76.2458 = 0.09965

Ka = [H+][OCl-]/[HOCl]

10^-7.53 = x*x/(M-x)

10^-7.53 = x*x/(0.09965)

x^2 =10^7.53 * 0.09965

x = sqrt(2.94087999e-9)

x = 0.00005423

ph = -log(0.00005423) = 4.266