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Chemistry

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A solution made up of 1.00 M NH3 and 0.500 M (NH4)2SO4 has a pH of 9.26.

a. Write the net ionic equation that represents the reaction of this solution with a strong acid.

b. Write the net ionic equation that represents the reaction of this solution with a strong base.

c. To 100. mL of this solution, 10.0 mL of 1.00 M HCl is added. How many moles of NH3 and NH4+ are present in the reaction system before and after the addition of the HCl? What is the pH of the resulting solution?

  • Chemistry - ,

    Jason, have you worked with the Henderson-Hasselbalch equation?

  • Chemistry - ,

    pH=pKa+ log(conjugate base/conjugate acid)


    I have it in my notes, I just can't figure out the steps to solve this problem, or how it applies.

  • Chemistry - ,

    The base is NH3.
    The acid is NH4^+.
    a. With a strong acid it's the base that uses it; i.e.,
    NH3 + H^+ ==> NH4^+

    b. With a strong base it's the acid that uses it.
    NH4^+ + OH^- ==> NH3 + H2O

    c. So we start with 100 mL of the buffer.
    millimols NH3 = mL x M = 100 x 1M = 100
    mmols NH4^+ = mL x M = 100 x 0.5M x 2 = 100
    mmols HCl added = 10mL x 1M = 10.

    ........NH3 + H^+ ==> NH4^+
    I.......100...0........100
    add...........10...........
    C.......-10..-10........+10
    E........90....0........110

    pH = pKa + log (base)/(acid_
    pH = pKa + log(90/110)
    Plug in pKa and solve for pH.

  • Chemistry - ,

    Ok, thank you! this makes a lot of sense. and the pKa is the -logKa? Ka=[NH4^+]=110?

  • Chemistry - ,

    Not quite.
    If you use Kb for NH3 = 1.8E-5 then pKb = -log Kb = about 4.74 and since
    pKa + pKb = pKw = 14, then
    pKa = 14-4.74 = 9.26. Your tables may give a different value for Kb NH3 but most show 1.8E-5 or 1.75E-5

  • Chemistry - ,

    what PH would mark the end-point of a weak acid having a ka value of 0.000005 assume the salt formed to be at a molar concentration of 0.05 at the end

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