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HNO3 with pH 3 is mixed with 2 L KOH with pH 12. assuming volumes are additive what is the pH of the final solution?

I have absolutely no idea how to do this. I do know the answer, though. I have a test tomorrow and I would really appreciate it if someone could help!!!!


    Why didn't you share the answer? Then we would know if I did it right or not?
    However, I don't think you copied all of the problem. HOW MUCH HNO3 with a pH of 3. It makes a huge difference.
    Here is how you do it.
    mL HNO3 x M = millimoles HNO3. (pH 3 makes M = 1 x 10^-3) = millimoles H^+..
    mL KOH x M = millimoles KOH = 2000 x 1 x 10^-2 (pH = 12, pOH = 2; therefore, OH^- = 1 x 10^-2 M).

    So which millimoles is the greater, the H^+ or the OH^-.
    Subtract the smaller from the larger divide the difference by TOTAL mL and that will be either the H^+ or the OH^- and you can get pH or pOH from that.


    oh, sorry!
    its 2 liters HNO3 and the answer is 11.65

    so it would be:
    -log [(20 - 2e-8) / 4000] = 2.3 = [OH]
    14 - 2.3 = 11.65 (without the rounding of the 2.3)

    why, though, would you subtract it from the other?


    HNO3 + KOH ==> KNO3 + H2O

    moles HNO3 = 2000 x 1 x 10^-3 = 2 mmoles.
    moles KOH = 2ooo x 1 x 10^-2 = 20 mmoles.

    Why subtract. This is an acid/base reaction, they neutralize each other. ALL of the HNO3 is GONE and you are left with an excess of 20-2 =18 mmoles KOH.
    So (OH^-) = 18 mmoles/4000 mL = 0.00450 M.
    pOH = 2.35 and pH = 11.65. :-).

    Isn't chemistry FUN?!

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