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Suppose you wanted to produce an aqueous solution of = 8.65 by dissolving the following salt KNO2. At what molarity would you use?

  • chemistry - ,

    Suppose you wanted to produce an aqueous solution of = 8.65 WHAT by dissolving the following salt KNO2. At what molarity would you use?

  • chemistry - ,

    I assume the desired pH is 8.65 although you did not label it, as DrBob pointed out.
    Ka = [H+][NO2-] / [HNO2]
    [H+] = antilog(-8.65)=2.24x10^-9 (a basic solution)
    [OH-] = Kw / [H+]
    [OH-] = (1x10^-14)/(2.24x10^-9) = 4.47x10^(-6)
    Ka = [H+][NO2-] / [HNO2] = 4.5x10^-4 (looked up)
    Kb = Kw / Ka = (1x10^-14) / (4.5x10^-4) = 2.22x10^-11
    The basic solution is formed by the hydrolysis of NO2-. The equilibrium constant is the same as the Kb:
    NO2-(aq) + H2O <=> HNO2 + OH-(aq)
    Kb = [HNO2][OH-] / [NO2-]
    assuming [HNO2]=[OH-],
    2.22x10^-11 = [4.47x10^(-6)][4.47x10^(-6)] / {[NO2-]-[4.47x10^(-6)]} or.
    2.22x10^-11 = [4.47x10^(-6)][4.47x10^(-6)] / [NO2-], since 4.47x10^(-6) is very small compared to [NO2-].
    Solve for [NO2-] which is the same as the concentration of NaNO2.

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