Solve Acid-Base Equilibrium: Calculate Ka of HA

[likeachild]

Homework Statement

1) A 0.682-gram sample of an unknown weak monoprotic organic acid, HA, was dissolved in sufficient water to make 50 milliliters of solution and was titrated with a 0.135-molar NaOH solution. After the addition of 10.6 milliliters of base, a pH of 5.65 was recorded. The equivalence point (end point) was reached after the addition of 27.4 milliliters of the 0.135-molar NaOH.

(a) Calculate the number of moles of acid in the original sample.
Done. 3.70 * 10^-3 mol HA
(b) Calculate the molecular weight of the acid HA.
Done. 184.37 g/mol
(c) Calc the number of moles of unreacted HA remaining in solution when the pH was 5.65.
done. 0.0027 mol
(d) Calculate the [H3O+] at pH = 5.65.
done. 2.2 * 10^-6
(e) Calculate the value of the ionization constant, Ka, of the acid HA.
I am using an ICE chart, with the equation HA <-> H+ + A-
The initial concentration for HA I used was 0.0027 mol/ 0.050 L
the initial concentration for H+ was 2.2 * 10 ^-6, which is also equal to the concentration of A-.
However, when I plug in these values into the equilibrium expression, Ka = ([H+][A-])/[HA], I get 1.1 * 10^ -10, and the answer key says that the correct answer is 1.4 * 10 ^-6.

Am I doing something wrong here, or is the answer key wrong?

 

[chemisttree]

Show your work on e). What values of [H+], [A-] and [HA] did you use?

 

[likeachild]

Ka = $$\frac{[H+][A-]}{[HA]}$$

[H+] = 2.2 * 10^-6 (got this from pH 5.65)
[A-] = 2.2 * 10^-6
[HA] = 0.0027 mol/ 0.050 L

 

[Kalirren]

If you write out all of the equations you have and plug them into each other, you'll see that [H+] is not equal to [A-]. Your assumption that they are equal is throwing you off.

Also don't forget that the volume of the entire solution changed because you added the NaOH solution.

 

[likeachild]

thanks.
I forgot about all the [A-] ions that came into the system as a result of the neutralization of the weak acid with NaOH.