Answer to Question #202993 in General Chemistry for Mickayla

Question

An acid HA has K_a = 2.28 x 10 ^-4. The % ionisation of this acid in a 0.170 M solution of the acid in water is closest to

1. 2.930 %
2. 3.662 %
3. 2.564 %
4. 0.498 %
5. 0.436 %
6. 0.623 %

Answer 1

Since this is a weak acid, the reaction is as follows: HA + H20 <==> A- + H3O+

Recall that percent ionization = (Concentration of the hydronium ion [H3O+] at equilibrium divided by the initial concentration of the weak monoprotic acid) x 100

Given: Initial concentration of HA = 0.153M

To find the [H3O+], create an ICE table.

HA H20 A- H3O+

Initial 0.153 - 0 0

Change -x +x +x

Equilibrium 0.153 - x x x

Now write out the formula for the acid ionization constant Ka,

where Ka = ( [A-]x [H3O+] )/ [HA] = x2 / 0.153 - x = 4.43 x 10-3

5)0.436 %