Anonymous
Anonymous asked in Science & MathematicsChemistry · 9 years ago

Chemistry Question dealing with pH?

Calculate the Ka of a solution of 0.00250 M of H2PO4^-1 with a pH of 4.90.

H2P04^-2 (aq) + H2O (l) <----> H3O^+ + HPO4^-2

I don't know if the first part of the question (the H2PO4^-1 part) should be H2PO4^-2. Or if that matter to the problem. Please show me how you did this. I can't figure it out on my own. Help me please!

2 Answers

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  • 9 years ago
    Favorite Answer

    it must be H2PO4- + H2O <--> H3O+ + HPO42-

    [H+] = 10^-4.9

    [H+] = 1.2589 x 10^-5

    [H+] = V(Ka x M)

    1.2589 x 10^-5 = V{Ka x 0.00250) --> square this equation

    1.5849 x 10^-10 = Ka x 0.00250

    Ka = 6.3396 x 10^-8

    Hope this helps!

  • nasser
    Lv 4
    4 years ago

    a million)i think of you advise what concentration of hydroxide neutralizes a pH of 5. thus we would desire to comprehend what number hydroxide ions are mandatory to realize a pH of 7. 10^-5M -[OH]=10^-7M [OH-]=9.9x10^-6M (i'd get 2nd evaluations in this) 2) a) HCN(aq) + H2O(l)--> H3O+(aq) + CN-(aq) b)Sr(OH)2(aq) + H2O(l) --> Sr2+(aq) + OH-(aq)

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