What is the Ka of a weak acid (HA) that produces a pH of 3.45 with 0.1M?
2.54E-3
3.45E-3
1.27E-6
3.56E-3
5.48E-4
pH = 3.45
3.45 = -log(H^+)
(H^+) = approx 3.55E-4.
...........HA ==> H^+ + A^-
I.........0.1.....0.....0
C..........-x.....x.....x
E.........0.1-x...x.....x
where x = 3.55E-4
Substitute that into Ka expression and solve for Ka.
To find the Ka of a weak acid (HA) that produces a pH of 3.45 with a concentration of 0.1M, we can use the equation for the dissociation of the weak acid:
HA ⇌ H+ + A-
The Ka expression for this reaction is:
Ka = [H+][A-] / [HA]
In this case, [H+] represents the concentration of H+ ions, which can be calculated from the pH using the equation:
[H+] = 10^(-pH)
Substituting the given pH value of 3.45 into the equation, we get:
[H+] = 10^(-3.45)
[H+] ≈ 3.56E-4 M
Since the weak acid (HA) has a concentration of 0.1 M, [HA] = 0.1 M.
Substituting the values into the Ka expression, we have:
Ka ≈ (3.56E-4)(3.56E-4) / 0.1
Ka ≈ 1.27E-8
Therefore, the Ka of the weak acid (HA) is approximately 1.27E-8. None of the options provided exactly match this value, so it seems there is an error in the options.