The PH of a 1.3M HF solution is 3.18. What % of the acid ionized?
pH = -log(H^+)
3.18 = -log(H^+)
(H^+) = 6.6E-4 but you should confirm that.
...........HF ==> H^+ + F^-
...........1.3..6.6E-4..6.6E-4
%ionization = (6.6E-4/1.3)*100 = ?
thank you!!
To find the percentage of the acid ionized in a solution, we can use the concept of the acid dissociation constant (Ka) and the Henderson-Hasselbalch equation.
The Henderson-Hasselbalch equation is given by: pH = pKa + log([A-]/[HA])
In this case, HF is a weak acid and partially ionizes in water to form H+ and F- ions. The dissociation reaction for HF can be written as: HF ↔ H+ + F-
The Ka expression for this reaction is: Ka = [H+][F-]/[HF]
Since we are given the pH of the solution (pH = 3.18), we can convert it back to the [H+] concentration using the formula: [H+] = 10^(-pH)
Now, we need to find the concentration of [HF], [H+], and [F-]. Since the initial concentration of HF is given as 1.3M, [HF] = 1.3M.
To solve for the concentration of [H+], we substitute the given pH value into the formula: [H+] = 10^(-3.18) = 6.31 x 10^(-4) M
Using the equation Ka = [H+][F-]/[HF], we can substitute the known values:
Ka = (6.31 x 10^(-4) M) * (x) / (1.3 M)
where x represents the concentration of [F-].
Rearranging the equation, we get: x = (Ka * [HF]) / [H+]
x = (6.31 x 10^(-4) M) * (1.3 M) / (6.31 x 10^(-4) M)
x = 1.3 M
Now, the percentage of the acid ionized can be determined by dividing the concentration of [F-] by the initial [HF] concentration and multiplying by 100:
Percentage of acid ionized = ([F-] / [HF]) x 100
Percentage of acid ionized = (1.3 M / 1.3 M) x 100
Percentage of acid ionized = 100%
Therefore, 100% of the HF acid is ionized in the 1.3M HF solution.