Enough of a monoprotic acid is dissolved in water to produce a 0.0195 M solution. The pH of the resulting solution is 6.24. Calculate the Ka for the acid.
Also, the answer is not 1.7 * 10 ^ -11
The response is: You treated the initial concentration as 0. In this case, the initial 10^-7 M H+ from water is not negligible. If [H+]initial = 10 ^ -7 and [H+]final = 5.8 * 10^-7, by how much did the concentration change? What does that say about how much A- was produced?
nevermind, I figured it out
you figured it out then what was the solution how did you solve it then because others are having the same problems and the same errors
To calculate the Ka for the acid, we need to determine the concentration of the acid dissociated (A-) and the concentration of the undissociated acid (HA) in the solution.
We know that the pH of the solution is 6.24, which means that [H+] (concentration of H+ ions) in the solution is 10^(-pH). Thus, [H+] = 10^(-6.24) = 4.0 * 10^(-7) M.
Since the acid is monoprotic, the initial concentration of A- is equal to the concentration of H+ ions, which is 4.0 * 10^(-7) M. The concentration of undissociated acid (HA) can be determined by subtracting the change in concentration from the initial concentration.
The change in concentration of H+ ions can be calculated by subtracting the initial concentration of H+ ions (10^(-7) M) from the final concentration of H+ ions (4.0 * 10^(-7) M).
Change in [H+] = [H+]final - [H+]initial = (4.0 * 10^(-7)) - (10^(-7)) = 3.0 * 10^(-7) M
Since the molar ratio of A- to H+ is 1:1, the change in concentration of A- is also 3.0 * 10^(-7) M.
Therefore, the concentration of undissociated acid (HA) is the initial concentration of A- minus the change in concentration of A-:
[HA] = [A-] initial - [A-] change = (4.0 * 10^(-7)) - (3.0 * 10^(-7)) = 1.0 * 10^(-7) M
Now, we can use the equation for the dissociation of the acid:
HA ⇌ H+ + A-
Ka = ([H+][A-])/[HA]
Substituting the values we found, we get:
Ka = ((4.0 * 10^(-7)) * (3.0 * 10^(-7))) / (1.0 * 10^(-7)) = 12.0 * 10^(-14) or 1.2 * 10^(-13)
Therefore, the Ka for the acid is approximately 1.2 * 10^(-13).