Question

Find the percent ionization of a 0.100 M solution of a weak monoprotic acid having Ka= 1.2×10⁻³.

Express your answer using two significant figures.

Answer 1

Given:

[Monoprotic acid] = 0.100 M, ka = 1.2 E-3

Solution:

Assume HA and set up ICE chart,

            HA (aq) + H2O (l) --- > A^- (aq) + H₃O⁺(aq)

I           0.100                               0                  0

C         -x                                     +x                +x

E          (0.100-x)                     x                      x

Set up Ka expression.

Ka = [A^- ] [ H₃O⁺] / [HA]

Plug in equilibrium concentration and ka value.

1.2 E-3 = x²/(0.100 -x)

Simplify above equation.

1.2 E-3 x (0.100-x) =x²

1.2E-4 – 1.2 E-3 x = x²

X² + 1.2 E-3 x – 1.2 E-4 = 0

Use quadratic equation formula,

x =(- b +/- sqrt (b²-4ac) )/ 2a

Value of a , b and c

a = 1, b = 1.2 E-3 , c =- 1.2 E-4

Plug in these value to get the value of x.

x = (-(1.2E-3)+sqrt ((1.2E-3)²-4 x 1 x (-1.2 E-4)) / 2 x 1

= 0.01037

[H₃O^­+]=x = 0.01037 M

Percent ionization =( x / 0.1 ) x 100

Percent ionization = 0.01037 / 0.1 ) x 100

= 10.4 %

Percent ionization of this given acid would be 10.4 %