For many weak acid or weak base calculations, you can use a simplifying assumption to avoid solving quadratic equations. Classify these situations by whether the assumption is valid or the quadratic formula is required.

a. [HA]=0.01M Ka=1x10^ - 4:

b. [HA]=0.01M Ka=1x10^ - 5:

c. [HA]=0.1M Ka=1x10^ - 3:

d.[HA]=1M Ka=1x10^ - 3:

e. [HA]=0.001M Ka=1x10^ - 5

a. not valid

b. valid

c. not valid

d. valid

e. not valid

Explanation:

The assumption to avoid solving the quadratic equation for the calculation of [H⁺] and [OH⁻] involved in the equilibria of weak acids and bases (small Ka and Kb) is valid as long as the value obtained from the shortcut is less than 5 % or less of the original acid or base concentration.

For a general monoprotic acid, as in this question, the equlibria is:

HA + H₂O ⇄ H₃O⁺ + A⁻ Ka = [H₃O⁺][A⁻]/[HA]

To determine the concentrations at equilibrium we are going to setupup the ICE table:

[HA] [H₃O] [A⁻]

Initial [HA]₀ 0 0

Change - x + x + x

Equil [HA]₀ - x x x

Ka = x² / [HA]₀ - x

Here is where we make our simplification of approximating [HA]₀ - x to the original acid concentration, [HA]₀, assuming x is much less than [HA] since HA is a weak acid.

To answer our questions we will solve for x, and then can compare it to the initial HA concentration.

Lets now perform our calculations.

(a) x = √ (0.01 x 1x 10⁻⁴) = 1 x 10⁻³ M = [H₃O⁺]

% = 1 x 10⁻³/.01 x 100 = 10%

The assumption is not valid.

(b) x = √ (0.01 x 1x 10⁻⁵) = 3.2 x 10⁻⁴ M = [H₃O⁺]

% = 3.2 x 10⁻⁴ / 0.01 x 100 = 3.2 %

The assumption is valid since the criteria of 5 % or less has been met.

(c) x = √ (0.1 x 1x 10⁻³) = 1.0 x 10⁻² M = [H₃O⁺]

% = 1.0 x 10⁻² / 0.1 x 100 = 10 %

The assumption is not valid, we wiould have to solve the quadratic equation.

(d) x = √ (1 x 1x 10⁻³) = 3.2 x 10⁻² M = [H₃O⁺]

% = 3.2 x 10⁻² / 1 x 100 = 3.2

The assumption is valid.

(e) x = √ (0.001 x 1x 10⁻⁵) = 1.0 x 10⁻⁴ M = [H₃O⁺]

% = 1.0 x 10⁻⁴ /.001 = 10 %

The assumption is not valid and one has to solve the quadratic equation.