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March 25, 2017

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What mass of solid potassium nitrite would need to be dissolved in 500. mL of 0.40 M HNO2 to make 500. mL of a NO2−/HNO2 buffer solution that has a pH = 3.25? Ka of HNO2 = 7.1 x 10−4. (Assume no significant volume change during dissolution of the potassium nitrite)
(1) 3.3 g (2) 6.5 g (3) 11 g (4) 21 g (5) 42 g

I know the answer is 21 g, but I cannot seem to get this answer when I'm working out the problem.

  • Chemistry - ,

    500 mL x 0.40M = 200 millimols HNO2 = 0.200 mol HNO2
    3.25 = 3.15 + log(base)/(.200)
    mols base = 0.252
    0.252 x 85.1 = 21.4 g.

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