Asked by marie
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.
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.
Answers
Answered by
DrBob222
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.
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.
Answered by
marie
oh! i get it now, thank you very much =)
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