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.
The value of the percent dissociation of the acid determines whether the quadratic equation will apply or not.
The acid dissociation constant gives the extent of dissociation of an acid in solution.
We may not solve a quadratic equation when the hydrogen ion concentration is less than 5 %.
Given a hypothethical monoprotic acid:
HA + H₂O ⇄ H₃O⁺ + A⁻ Ka = [H₃O⁺][A⁻]/[HA]
From the ICE table:
HA H₃O^+ A⁻
I HA 0 0
C - x +x +x
E HA - x x x
Ka = x² / HA - x
Lut us now find x^2 in each case;
(a) x = √ (0.01 x 1x 10⁻⁴) = 1 x 10⁻³ M = [H₃O⁺]
% = 1 x 10⁻³/.01 x 100 = 10%
This assumption is not valid here since we obtained 10%.
(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 %
This assumption is valid here since we obtained 3.2 %.
(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 %
This assumption is not valid here since we obtained 10%.
(d) x = √ (1 x 1x 10⁻³) = 3.2 x 10⁻² M = [H₃O⁺]
% = 3.2 x 10⁻² / 1 x 100 = 3.2%
This assumption is valid here since we obtained 3.2 %.
(e) x = √ (0.001 x 1x 10⁻⁵) =1.0 x 10⁻⁴ M = [H₃O⁺]
% = 1.0 x 10⁻⁴ / .001 = 10 %
This assumption is not valid here since we obtained 10%.
Learn more about acid dissociation: https://brainly.com/question/1298425
