Anonymous
Anonymous asked in Science & MathematicsChemistry · 6 months ago

If a 100 mL sample of sodium oxalate (Na2OOCCOO (aq)) has a pH of 9.23 at 25.0 °C, the initial concentration of the sample of sodium ?

sodium oxalate is?

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  • 6 months ago
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    Oxalic acid has two ionizable carboxyl groups. The Ka's for the two are 5.4X10^-2 and 5.4X10^-5. In a solution of sodium oxalate, only the second of these will be involved in a basic reaction. That reaction is:

    C2O42- + H2O <--> HC2O4- + OH-

    The expression for this Kb is:

    Kb = [HC2O4-][OH-]/[C2O42-]

    The Kb for that group will be:

    Kb = 1X10^-14 / 5.4X10^-5 = 1.85X10^-10

    Now, in that solution, [OH-] = [HC2O4-]

    [OH-] can be calculated from the pH as:

    pOH = 14 - 9.23 = 4.77

    [OH-] = 10^-pOH = 1.7X10^-5 M = [HC2O4-]. So:

    Kb = 1.85X10^-10 = (1.7X10^-5)^2 / [C2O42-]

    [C2O42-] = 1.6 M = initial concentration of sodium oxalate

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