Answer:
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.
