CHEMISTRY

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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...

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