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Determine pH of each solution:

0.20 M KI

i don't know how you know if you use..
K+H2O->KOH + H
I- +H2O--> HIO +H

Im thinking this one...

I- + H2O --> HIO + H

there is a Ka value for HIO so that is why i made it a product...right? Ka HIO= 2.3X10^2


2.3X10-11 = x^2/ 0.20-x

then do pH.. would that be correct?
so pH=5.7 ?

  • CHEMISTRY!!!=0 -

    You don't use either because neither K^+ nor K^- hydrolyze. You can TRY to write the equations like this;
    K^+ + HOH ==> KOH + H^+ but when you go to look up Kb for KOH you realize it has no Kb because it is 100% ionized.
    Then I^- + HOH ==> HI + OH^- and you look up Ka for HI but you quickly realize HI has no Ka because it is a strong acid (100% ionized). Therefore, when KI dissolves in water it is the salt of a strong acid (HI) and a strong base (KOH), neither has a Ka nor Kb so you have the salt dissolved in water and the pH = 7.0 for pure water. Technically, it won't be quite 7.0 because of activity coefficients but you haven't studied that yet.
    Back to your former question, another good way to know is you look at the equation you just wrote with water; i.e.,
    CH3NH3^+ + H2O ==>H3O^+ + CH3NH2
    so you know to use Kb for the base sitting there on the product side.
    If you have a base like acetate, then
    C2H23O2^- + HOH ==> HC2H3O2 + OH^- and you know to look up the Ka value for HC2H3O2, acetic acid, since that's the acid sitting there.

  • CHEMISTRY!!!=0 -

    OH, okay so when I produce a strong acid and a strong base it is neutral so pH=7. Meaning the 0.20M KI doesn't even matter?

  • CHEMISTRY!!!=0 -

    That's right. At least for now since you haven't studied the effects of activity coefficients.

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