Answer to Question #202993 in General Chemistry for Mickayla

Question #202993

An acid HA has Ka = 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 %
1
Expert's answer
2021-06-07T03:12:36-0400

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 %

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