Answer:
pH = 13
Step-by-step explanation:
Dissociation constant, Ka, of an acid HA in water, is written as:
HA(aq) + H₂O(l) ⇄ H₃O⁺(aq) + A⁻(aq)
Ka = [H₃O⁺][A⁻] / [HA] = 1x10⁻²⁰
Now, knowing the equilibrium of water is:
2H₂O ⇄ OH⁻ + H₃O⁺
Kw = [OH⁻] [H₃O⁺] = 1x10⁻¹⁴
Now, subtracting the equilibrium of water - the dissociation of the acid:
A⁻(aq) + H₂O ⇄ OH⁻ + HA
And the equilibrium constant, Kb, is:
Kb = Kw / Ka = 1x10⁻¹⁴ / 1x10⁻²⁰ = 1x10⁶
And is defined as:
Kb = 1x10⁶ = [OH⁻] [HA] / [A⁻]
If you make a solution of 0.1M NaA (A⁻ in water), the equilibrium concentrations are:
[A⁻] = 0.1M - X
[OH⁻] = X
[HA] = X
Where X is the reaction coordinate
Replacing in Kb formula:
1x10⁶ = [X] [X] / [0.1-X]
1x10⁵ - 1x10⁶X = X²
0 = X² + 1x10⁶X - 1x10⁵
Solving for X:
X = -1 → Wrong solution. There is no negative concentrations
X = 0.09999999M → Right solution.
As [OH⁻] = X, [OH⁻] = 0.09999999M,
Defining pOH = -log [OH⁻]
pOH = 1
As pH = 14- pOH
pH = 14 - 1
pH = 13