# CHEMISTRY

posted by on .

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?

• CHEMISTRY - ,

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.

• CHEMISTRY - ,

Uh how do you convert the 5.25% to molarity?

• CHEMISTRY - ,

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

• CHEMISTRY - ,

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

• CHEMISTRY - ,

What gets substituted into the formula M1V1=M2V2? I don't understand...