The pKa of hypochlorous acid is 7.530. A 54.0 mL solution of 0.113 M sodium hypo

Question

The pKa of hypochlorous acid is 7.530. A 54.0 mL solution of 0.113 M sodium hypochlorite (NaOCl) is titrated with 0.309 M HCl. Calculate the pH of the solution

a) after the addition of 5.92 mL of 0.309 M HCl.

b) after the addition of 21.0 mL of 0.309 M HCl.

The pKa of hypochlorous acid is 7.530. A 54.0 mL solution of 0.113 M sodium hypochlorite (NaOCI) is titrated with 0.309 M HCI. Calculate the pH of the solution a) after the addition of 5.92 mL of 0.309 M HC Number pH- b) after the addition of 21.0 mL of 0.309 M HCI. Number c) at the equivalence point with 0.309 M HC Number PH-

Explanation / Answer

a)

mmol of OCl = MV = 54*0.113 = 6.102

mmol of H+ = MV = 0.309*2.92 = 0.90228

then..

mmol of OCl left =6.102- 0.90228 = 5.19972

mmol of HOCl = 0.90228

pH = pKa + log(OCl-/HOCl)

pH = 7.53 + log(5.19972/0.90228)

pH = 8.290

b)

mmol of OCl = MV = 54*0.113 = 6.102

mmol of H+ = MV = 0.309*21 = 6.489

mmol of HOCl formed = 6.102

mmol of H+ left = 6.489- 6.102 = 0.387

Since H+ is stronger than HOCl, then expect

pH = -log([H+])

[H+] = mmol left / V total = 0.387 / ( 21+54) = 0.00516

pH = -log(0.00516) = 2.287

c)

in equivalence

6.102 mmol of OCl- form --> HOCl

then

[HOCl] = 6.102 / (Veq + Vacid)

Veq = 6.102 / (0.309) = 19.74 mL

[HOCl] = 6.102 / (19.74 + 54)

[HOCl] = 0.082750

then, there is ioniaztion so

HOCl <--> H+ + OCl-

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

10^(-7.53) = x*x/(0.082750 -x)

x = 4.94*10^-5

pH = -log(x) = -log( 4.94*10^-5

pH = 4.31

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