Posted by TYLER CHIRECE on Wednesday, February 16, 2011 at 12:44am.
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...
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