Question

Question 22 of 25 Submit An unknown weak acid with a concentration of 0.071 M has a pH of 1.80. What is the Ka of the weak acid?

Answer 1

Let's assume, HA is the weak acid for the sake of this question. The dissociation of HA would look something like this:

HA+H,0 + H30++ A-

Since this is a weak acid, it wouldn't dissociate completely in water. So, in order to arrive at an algebraic expression for Ka, we would write an ICE table for this reaction:

	HA	H3O+	A-
Initial concentration	0.071 M	0	0
Change in concentration	-x	+x	+x
Equilibrium concentration	(0.071-x) M	x	x

Now, we can write the expression for Ka:

K = [H3O+][A-] [HA] (0.071 - x)

Here, since the pH is given, we can ascertain the concentration of hydronium ion which is equal to x.

[+o®H]601- = Hd

1.80 = -log(x)

= 0.0158 M

Plugging the value of x into the relation for Ka:

K = 70.071 – 0. (0.0158) (0.071 -0.0158) = 4.52 x 10-3