Asolution of household bleach contains 5.25% sodium hypochlorite, naocl, by mass. assuming that the density of bleach is the same as water, calculate the volume of household bleach that should be diluted with water to make 500.0 ml of a ph = 10.18 solution. use the ka of hypochlorous acid found in the chempendix.

Let V ml of household bleach is use.
Weight percent is 5.25 and density is 1 g / ml
The mass of NaOCl = V ml x 1 g/ml x (5.25/100) = 0.0525 V g
Molar mass of NaOCl is 74.4 g/mol
Total volume is 500.0 ml or 0.5 L
The concentration of NaOCl in 500 ml solution:
C = (0.0525 V g / 74.4 g/mol) / 0.500 L = 0.001411 V M --> (1)
pH is 10.18
For HOCl Ka = 3.5 x 10⁻⁸ 
pKa = - log Ka = - log (3.5 x 10⁻⁸) = 7.46
For a salt of weak acid and strong base:
pH = 7 + 0.5 pKa + 0.5 log C
10.18 = 7 + 0.5 (7.46) + 0.5 log C
log C = - 1.096
C = 0.080 M --> (2)
0.080 M = 0.001411 V M
V = 57 mL

The answer is c) pyrite