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.
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.
2 answers
oh! i get it now, thank you very much =)
1 answer