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.