Consider an acid, HA, with pKa = 8.259. 0.129 moles of hydroxide ion are added to 13.83mL of a 0.931M solution of the acid. What is the resulting pH?
So I know pH = pKa + log([acid]/[base])
So really, how would I find the concentration of the base and the new concentration of the acid? If I have those, I can do the rest.
Thanks for any help!
2006-10-23 16:11:42 · 2 answers · asked by Anonymous in Science & Mathematics ➔ Chemistry
6.264 is the answer, but I seem to not be able to get that.
2006-10-23 16:38:57 · update #1
The moles of acid = 0.01383L x 0.931M = 0.0129 mol HA
The moles of base = 0.129 mol OH-
The excess moles of base = 0.116 mol OH-
Since the question does not state that the volume changes upon addition of the OH-, one must assume the same volume at the end of the addition = 0.01383L
The concentration of the excess OH- = 0.166 mol OH- / 0.01383L = 8.39M OH-
pOH = -log(8.39) = -0.924
pH = 14 – (-0.924) = 14.924
2006-10-23 16:25:34 · answer #1 · answered by Ravenwoodman 3 · 0⤊ 2⤋
Henderson equation: pH = pKa + log([base]/[acid]) solve for pKa
2016-03-28 05:41:47 · answer #2 · answered by Anonymous · 0⤊ 0⤋