20. NEED HELP FAST

Question

The  p𝐾a  of hypochlorous acid is  7.530 . A  52.0  mL solution of  0.145  M sodium hypochlorite ( NaOCl ) is titrated with  0.321  M  HCl .

Calculate the pH of the solution after the addition of  7.23  mL of  0.321  M  HCl .

pH=

Calculate the pH of the solution after the addition of  24.6  mL of  0.321  M  HCl .

pH=

Calculate the pH of the solution at the equivalence point with  0.321  M  HCl .

pH=

Answer 1

Hypochlorous acid = HClO

A solution of HClO + NaOCl constitutes a buffer, and the pH can be determined from the Henderson Hasselbalch equation: pH = pKa + log [NaOCl]/[HClO]

a) moles NaOCl = 0.052 L x 0.145 mol/L = 0.00754 moles

moles HCl = 0.00723 L x 0.321 mol/L = 0.00232 moles

a) NaOCl + HCl ==> HOCl + NaCl

0.00754......0.00232...0................Initial

-0.00232....-0.00232......+0.00232 ...Change

0.00522..........0.........0.00232........Equilibrium

pH = 7.53 + log

pH = 7.53 + log (0.00522/0.00232)

pH = 7.88

b) NaOCl + HCl ==> HOCl + NaCl

0.00754.....0.00790.......0.................Initial

-0.00754....-0.00754........+0.00754.......Change

0...............0.00036........0.00754....Equilibrium

Final volume = 52 ml + 24.6 ml = 76.6 ml = 0.0766 L

[H+] = 0.00036 mol H+/0.0766 L = 0.0047 M

pH = -log 0.0047 = 2.3 (neglecting any contribution of H+ from the 0.00754 moles HClO)

c) At the equivalence point, all of the NaOCl has been converted to HOCl

(x L)(0.321 mol/L) = 0.00754 moles and x = 0.0235 L = 23.5 mls

Final volume at equivalence = 52.0 ml + 23.5 ml = 75.5 mls = 0.0755 L

Final [HClO] = 0.00754 mol/0.0755 L = 0.0999 M

Ka = 1x10^-7.530 = 2.95x10^-8

Ka = 2.95x10^-8 = [H+][ClO-]/[HClO]

2.95x10^-8 = (x)(x)/0.0999

x² = 2.95x10^-9

x = [H+] = 5.4x10^-5 M

pH = -log 5.4x10^-5 = 4.27