Question

The pk, of hypochlorous acid is 7.530. A 58.0 mL solution of 0.110 M sodium hypochlorite (NaOCI) is titrated with 0.343 M HCI. Calculate the pH of the solution after the addition of 7.42 mL of 0.343 M HCI. pH = Calculate the pH of the solution after the addition of 19.7 mL of 0.343 M HCI. pH = Calculate the pH of the solution at the equivalence point with 0.343 M HCI. pH =

Answer 1

1) pH after addition of 7.42ml addition

Number of moles of ClO^- = (0.110mol/1000ml) × 58.0ml = 0.006380mol

Number of moles of HCl added = (0.343mol/1000ml)×7.42ml = 0.002545mol

HCl + ClO^- -------> HClO + Cl^-

0.002545moles of HCl reacts with 0.002545moles of ClO^- to give 0.002545moles of HClO

After addition of HCl

number of moles of OCl^- = 0.006380mol - 0.002545mol = 0.003835mol

number of moles of HClO formed = 0.002545mol

Total volume = 58.0ml + 7.42ml =65.42ml

[ClO^-] = (0.003835mol/65.42ml) × 1000ml = 0.05862M

[HClO] = (0.002545mol/65.42ml)×1000ml = 0.03890M

Henderson - Hasselbalch equation

pH = pKa + log([A^-] /[HA])

pH = 7.530 + log( 0.05862M/ 0.03890M)

pH = 7.530 + 0.178

pH = 7.71

2) pH after addition of 19.7ml addition

1:1 molar reaction , so

C1×M1 = C2× M2

V2 = C1×M1 / C2

= 0.110M × 58.0ml / 0.343M

= 18.60ml

Therefore,

equivalence point = 18.60 ml

exceess volume of HCl added = 19.7ml - 18.60ml = 1.1ml

moles of HCl = ( 0.343mol/1000ml) × 1.1ml = 0.0003773mol

Total volume = 58.0ml + 19.7ml = 77.7ml

[HCl ] =(0.0003773mol/77.7ml) × 1000ml = 0.004856M

[H⁺] = 0.004856M

pH = -log[H⁺]

pH = -log(0.004856M)

pH = 2.31

3) pH at equivalence point

At equivalence point all the ClO^- converted to HClO

Total volume = 58.0 ml + 18.60ml = 76.6ml

[HClO] = 0.110M/(76.6ml/58.0ml ) = 0.08329M

HClO partly dissociates

HClO <-----> H⁺ + ClO^-  

Ka = [H⁺] [ClO^-]/[HClO] = 2.95×10^-8

at equillibrium

[HClO] = 0.08329 - x

[H⁺] = x

[ClO^-] = x

so,

x²/( 0.08329 - x) = 2.95 ×10^-8

solving for x

x = 0.000049

[H⁺] = 0.000049M

pH = -log(0.000049M)

pH = 4.31