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chem

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what is the pH change after the addition of 10mL of 1.0M sodium hydroxide to 90mL of a 1.0M NH3/1.0M NH4^+ buffer? Kb ammonia= 1.8 x 10^-5

i found the pH for the original buffer to be 9.255, but i'm not sure how to go about the rest of the question, please, someone help me.

  • chem - ,

    I would do this.
    I assume you used the HH equation to arrive at the answer of 9.255 (with which I agree). At that time you substituted 1M each for base and acid and log 1 = 0 which makes pH = pKa. When you do that base/acid = 1 which means (base) = (acid) and the total is 1M.
    90mL x 1M = 90 millimoles total or
    45 mmols base and 45 mmoles acid.
    The equation for the addition of NaOH is
    ............NH4+ + OH^- ==> NH3 + H2O
    initial.....45.....0........45......0
    add OH-............10...............
    change.....-10....-10.....+10......+10
    equil.......35.....0........55......10

    pH = 9.255 + log(55/35) = ??
    I would round the final answer to 2 places.

  • chem - ,

    oh! i get it now, thank you very much =)

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