Question 13.11: Consider sodium hypochlorite, NaOCl, the main component in h...

Consider sodium hypochlorite, NaOCl, the main component in household bleach. The hypochlorite ion,OCl^-, has K_b = 3.6 \times 10^{-7}. A solution is prepared by dissolving 12.0 g of NaOCl (MM = 74.45 g/mol) in enough water to make 835 mL of solution.

a  What is the pH of the solution?

ANALYSIS

Information given:                   K_b for OCl^{-}( = 3.6 \times 10^{-7})                                                                                                                 mass of NaOCl (12.0 g); molar mass of NaOCl (74.45 g/mol)                                                                  volume of solution (0.835 L)
Information implied:                    K_{w}
Asked for:                                       pH of the solution

STRATEGY

1. Determine the original concentration,[\quad ]_o, of NaOCl.
2. Draw a table as illustrated in Example 13.5.
3. Write the K expression and find [OH^-] assuming [OH^-]_{eq} << [OCl^-]_o. Check the validity of the
assumption by finding % ionization.
4. Find [H^+] using Equation 13.1 and pH using Equation 13.3.

K_w = [H^+][OH^-]              (13.1)

pH = -\log_{10}[H^+] = -\log_{10}[{H_3O}^+]              (13.3)

b Household bleach is 5.25% NaOCl by mass. Assuming that its density is 1.00 g/mL, is household bleach more alkaline than the prepared solution?

ANALYSIS

Information given:             NaOCl content of household bleach (5.25% by mass)                                                                                 density of bleach (1.00 g/mL)
pH of solution in part (a) (10.41)
Asked for:                                 Compare pH of solution (a) and pH of bleach.

STRATEGY

1. Assume 100.0 g (= 100.0 mL) of bleach. Thus, there are 5.25 g of NaOCl in 100.0 mL of
solution.
2. Find [OH^-], [H^+], and pH of bleach as in part (a).
3. Compare the pH of both solutions. The solution with a higher pH is more alkaline.