If the pH = 5.4 then
5.4 = -log(H^+)
-5.4 = log(H^+)
(H^+) = 3.98 x 10^-6
I didn't work through all the numbers to see if they match up but it doesn't matter what they are. If the pH is 5.4, then (H^+) is defined by that and nothing else.
if the pH of a half-neutralized acid solution is 5.4, how would you find the [H+] of the the solution. There is 1.0 g of L-ascorbic acid which was dissolved in 100 ml of water. That solution was split in two and 50 ml of the solution was titrated with 0.2 M NaOH (13 ml NaOH). then the other half of the solution was added to get the pH of 5.4. With this information how can I get the concentration of H+?
4 answers
oh ok.. thanks so much... So when you have the pH of a substance the [H^+] is always equal to 10^-pH?
is this then equal to the Ka of the acid solution?
Yes to the first question. (H^+) = 10^-pH.
At the half way point, which is what you had, then Ka = (H^+ or pKa = pH but that is only true at the half way point. You can see for a weak acid, such as HA
HA ==> H^+ + A^- and
Ka(H^+)(A^-)/(HA). Solving for (H^+) we get
(H^+) = Ka(HA)/(A^-). SO, when exactly half the HA has been neutralized, then there has been formed an equal amount of (A^-) so (HA)=(A^-) and (H^+) = Ka. You can take the -log of each side to obtain
-log(H^+) = -log Ka AND
pH = pKa. But remember, for this second part, this is true ONLY at the half way point to neutralization of the acid (Or, of course, when you make a buffer solution containing equal amounts of acid and salt).
At the half way point, which is what you had, then Ka = (H^+ or pKa = pH but that is only true at the half way point. You can see for a weak acid, such as HA
HA ==> H^+ + A^- and
Ka(H^+)(A^-)/(HA). Solving for (H^+) we get
(H^+) = Ka(HA)/(A^-). SO, when exactly half the HA has been neutralized, then there has been formed an equal amount of (A^-) so (HA)=(A^-) and (H^+) = Ka. You can take the -log of each side to obtain
-log(H^+) = -log Ka AND
pH = pKa. But remember, for this second part, this is true ONLY at the half way point to neutralization of the acid (Or, of course, when you make a buffer solution containing equal amounts of acid and salt).