The pH of an acid solution is 5.90. Calculate the Ka for the monoprotic acid. The initial acid concentration is 0.10 M.
What I have done:
pH = 5.9
a = 10^-5.9 = 1.2589 * 10^-6
Ka = [H+][A-]/[HA]
Ka = (1.2589*10^-6)^2/(.01-1.2589)
I got -1.26903013E-12, but it's not accepted as a correct answer.
Ka = (1.2589*10^-6)^2/(.1-1.2589E-6) was your error, so you can approximate this with
(1.2589*10^-6)^2/(.1 ) =
= 1.58 e-11
Your approach is correct, but there seems to be a mistake in your calculation. Let's go through the steps again:
1. Given that pH = 5.90, we can use the equation pH = -log[H+].
Therefore, 5.90 = -log[H+].
2. Rearranging the equation, we have [H+] = 10^(-pH).
[H+] = 10^(-5.90) = 1.2589 x 10^(-6).
3. Since the acid is monoprotic, the concentration of HA is equal to the initial acid concentration (0.10 M).
[HA] = 0.10.
4. To find the concentration of A-, we need to subtract the concentration of H+ from the initial acid concentration:
[A-] = [HA] - [H+] = 0.10 - 1.2589 x 10^(-6).
5. Now we can calculate the Ka using the equation Ka = [H+][A-]/[HA].
Ka = (1.2589 x 10^(-6))^2 / (0.10 - 1.2589 x 10^(-6)).
Calculating this expression, we find that Ka is approximately 2.006 x 10^(-12). Therefore, the correct answer is 2.006 x 10^(-12).