CHEMISTRY
posted by TYLER CHIRECE .
A solution household bleach contains 5.25% NaClO, by mass. Assuming that the density of bleach is the same as water(1.0g/ml). Calculate the volume of household bleach that should be diluted with water to make 500.0 ml of a ph= 10.26 solution.
So, the first thing i do was to find kb using the ka of hypochlorous acid but from there i don't know where to go. Could you help me?

You are right to calculate Kb.
ClO^ + HOH ==> HClO + OH^
Set up an ICE chart. You will substitute as follows:
Kb = (Kw/Ka) = (HClO)(OH^)/(ClO^)
You know Kw and Ka. (HClO)=(OH^) = x and (ClO^) = 5.25% BUT that must be converted to molarity. Solve for x which is the OH^, convert that to pOH, then to pH, then to (H^+). That is the (H^+) of the 5.25% bleach. Then use the dilution formula of
mL x M = mL x M. 
Uh how do you convert the 5.25% to molarity?

wouldnt the answer be 500 ml? since the density of water is the same as the bleach?

Uh how do you convert the 5.25% to molarity?
if the solution is 5.25% then 1 litre (=1 kg) contains 52.5 g
Calculate the molecular mass for NaClO=M
then the molarity =
52.5 g/M g mole^1 
What gets substituted into the formula M1V1=M2V2? I don't understand...