When 20.0 mL of 0.10 M NaOH are added to 20.0 mL of 0.10 M hypobromous acid, HOBr, what is the hydrogen ion concentration (H^+)? Ka for HOBr is 2 x 10^-9.
I get 5 x 10^-4. Is that right?
I don't think so.
HOBr + NaOH ==> NaOBr + H2O
20 x 0.1 HOBr = 2 mmoles HOBr.
20 x 0.1 NaOH = 2 mmoles NaOH
Those two exactly neutralize each other; therefore, the pH is determined by the salt, NaOBr.
(NaOBr) = 2 mmoles/40 mL = 0.05 M.
It hydrolyzes as a base,
BrO^- + HOH ==> HOBr + OH^-
Set up ICE chart.
Kb = (Kw/Ka) = (x)(x)/0.05
solve for x = (OH^-) and convert to H^+.
To find the hydrogen ion concentration (H^+) in this solution, you can use the concept of acid dissociation and the expression for the acid dissociation constant (Ka).
First, let's write the balanced chemical equation for the reaction between NaOH and HOBr:
NaOH + HOBr -> H2O + NaOBr
Since both NaOH and HOBr are strong electrolytes, they completely dissociate in water. Therefore, we can write:
NaOH -> Na+ + OH-
HOBr -> H+ + OBr-
Now we can calculate the concentration of OH- ions using the initial concentration of NaOH:
0.10 M NaOH * (20.0 mL / 40.0 mL) = 0.05 M OH-
Since OH- and H+ ions combine to form water in a 1:1 ratio, we know that the concentration of H+ ions is also 0.05 M.
However, we still need to consider the contribution of H+ ions coming from the dissociation of HOBr. The acid dissociation constant (Ka) for HOBr is given as 2 x 10^-9. Since the concentration of the original HOBr is 0.10 M, we can assume that most of it will dissociate, leaving only a small fraction undissociated.
Using the equation for the acid dissociation constant, Ka = [H+][OBr-] / [HOBr], we can calculate the concentration of H+ ions contributed by the dissociation of HOBr.
[H+][OBr-] / [HOBr] = 2 x 10^-9
[H+](0.05 M) / (0.10 M) = 2 x 10^-9
[H+] = (2 x 10^-9) * (0.10 M) / (0.05 M)
[H+] = 4 x 10^-9 M
Therefore, the hydrogen ion concentration (H^+) in the solution is 4 x 10^-9 M.
It seems that your result of 5 x 10^-4 is not correct. Double-check your calculations to see where you may have made an error.