Question

You mix 100.0 mL of 0.20 M HBr and 50.0 mL of 0.40 M NaClO. What is the pH of the resulting solution? Ka(HClO) = 3.5 x 10^-8. Provide your answer with solution.

Answer 1

Balanced equation for the reaction;

2HBr + NaClO $\to$ Br2 + NaCl + H2O

Moles of HBr

If 0.2 moles were in 1000mL,

How many will be in 100mL

$=\dfrac{0.2 mol x 100mL}{1000mL} =$ 0.02 mol

Moles of NaClO that is required to completely react with HBr;

Mole ratio of HBr : NaClO is 2:1

Moles of NaClO $=\dfrac{1}{2}x0.02 = 0.01mol$

But moles of NaClO present were;

0.4 mol were in 1000mL

How many will be in 50mL

$=\dfrac{0.4mol x 50mL}{1000mL} = 0.02mol$

Thus moles of NaClO are in excess by 0.02 mol - 0.01 mol = 0.01 mol

Therefore we are actually determining the pH of 0.01mol NaClO in 150mL of solution (since all the other products are neutral)

Concentration of the remaining NaClO;

0.01 mol were in 150mL of solution (100+50)

How many will be in 1000mL

$=\dfrac{0.01mol x 1000mL}{150mL} = 0.0667 M$

We then determine pH of 0.0667M NaClO

The salt will ionize as; (since Na+ is neutral)

ClO^- + H₂O $\iff$ HClO + OH^- (Kb = 3.45 x 10^-7)

I 0.0667 0 0

C -X +X +X

E 0.0667-X X X

$Kb = \dfrac{[HClO][OH]}{[ClO]}$

$0.0667x3.45x10^-7 = \dfrac{X^2}{0.0667} x 0.0667$

$0.0667x3.45x10^-7 = X^2$

finding squareroot of both sides

$\surd0.0667x3.45x10^-7 = X$

X = 4.7958x10^-4 = [OH]

pOH = -Log [OH]

= -Log 4.7958x10^-4

= 3.319

Since pOH + pH = 14

pH = 14-pOH

=14-3.319

= 10.68

Answer: pH = 10.68