Sodium hypochlorite (NaOCl,74.44 g/mol)is a common ingredient in household bleach.What is the pH of a particular brand of bleach that has a density of 1.11 g/mL and is 5.00% NaOCl by mass? (pK_a of HOCl = 7.46)

Question

Quizplus · Accepted Answer

First, we need to calculate the molarity of NaOCl in the solution:
5.00% mass percent solution means 5.00 g NaOCl in 100 g of solution.
Moles of NaOCl in 100 g of solution = (5.00 g NaOCl / 74.44 g/mol) = 0.067 moles
Volume of 100 g solution = (100 g / 1.11 g/mL) = 90 mL = 0.09 L
Molarity of NaOCl = 0.067 moles / 0.09 L = 0.74 M

Next, we need to determine the concentration of HOCl in the solution. NaOCl undergoes hydrolysis in water to produce HOCl and NaOH:
NaOCl + H2O ↔ HOCl + NaOH
Given that the pKa of HOCl is 7.46, we can assume that at pH less than 7, most of the hypochlorite ions will be protonated to form HOCl. At pH greater than 7, most of the hypochlorite ions will exist in the OCl- form. Thus, we need to calculate the pH at which the concentration of HOCl equals the concentration of OCl-.

We can use an ICE table to determine the equilibrium concentration of HOCl and OCl-:
NaOCl + H2O ↔ HOCl + NaOH
Initial:            0.74 M                0         0          0
Change:         - x                       + x     + x        +x
Equilibrium: 0.74 - x                   x      x         x

The equilibrium constant expression for this reaction can be written as:
Kw/Ka = ([H+][OH-])/[HOCl]

At pH = 10.67, we can assume that [H+] = 1x10^-10 M and [OH-] = 1x10^-4 M
Substituting these values and the given pKa of HOCl in the equation above, we get:
(1.0 x 10^-14) / (3.52 x 10^-8) = (1x10^-10)(1x10^-4) / [HOCl]
[HOCl] = 2.93 x 10^-7 M

Since [HOCl] = [OCl-] (due to the assumption made earlier), we can calculate the pH using the Henderson-Hasselbalch equation:
pH = pKa + log([OCl-]/[HOCl])
pH = 7.46 + log(2.93 x 10^-7 / 0.74)
pH = 10.67

Therefore, the pH of the bleach solution is 10.67, which corresponds to the concentration of hypochlorite ions being approximately equal to the concentration of hypochlorous acid. The correct answer is B.

Sodium Hypochlorite (NaOCl,74