CHEMISTRY QUESTION HELP: TITRATION QUESTION PLEASE HELP ASAP?

Calculate the pH when 25.0 mL of 0.100 M NaOH has been titrated to a 25.0 mL of 0.100 M nitrous acid (HNO2)? Assume at SATP.  

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  • 1 month ago

    At this point in the titration, you are at the equivalence point, having added as many moles of NaOH as you initially had of HNO2. Thus, you now have a solution of NaNO2 with a molarity of 0.0500 M (because you added an equal volume of NaOH).

    The NO2- ion acts as a base by the equation:

    NO2- + H2O <--> HNO2 + OH-

    Kb = [HNO2][OH-]/[NO2-]

    The value of Kb can be calculated from the value of Ka of HNO2 by the relationship:

    Ka X Kb = Kw = 1.00X10^-14

    Kb = 1X10^-14 / 4.0X10^-4 = 2.5X10^-11

    Now, in the solution, let [OH-] = [HNO2] = x, and [NO2-] = 0.050 - x. Because Kb is small, x will be small compared to 0.050 and so can be ignored. Then,

    Kb = 2.5X10^-11 M = x^2 / 0.050

    x = [OH-] = 1.1X10^-6 M

    pOH = 5.95

    pH = 14.00 - pOH = 8.05

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