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How many moles of NaBrO should be adder to 1L of .05M HBrO to form a buffer with pH=8.80?

Ka=2.5x10^-9

How would I solve this?

Do I need to do an ICE table?

Thanks!!

1 Answer

Relevance
  • Dr.A
    Lv 7
    1 decade ago
    Favorite Answer

    pKa = - log Ka = 8.6

    pH = pKa + log [BrO-] / [HBrO]

    8.80 = 8.6 + log [BrO-] / 0.05

    10^ 0.2 = 1.6 = [BrO-] / 0.05

    [BrO-] = 0.080 M

    moles BrO- = 0.080 M x 1 L = 0.080

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