Answer:
[tex]pH=13[/tex]
Explanation:
Hello,
In this case, given the acid, we can suppose a simple dissociation as:
[tex]HA\rightleftharpoons H^+ + A^-[/tex]
Which occurs in aqueous phase, therefore, the law of mass action is written by:
[tex]Ka=\frac{[H^+][A^-]}{[HA]}[/tex]
That in terms of the change [tex]x[/tex] due to the reaction's extent we can write:
[tex]1x10^{-20}=\frac{x*x}{0.1M-x}[/tex]
But we prefer to compute the Kb due to its exceptional weakness:
[tex]Kb=\frac{Kw}{Ka}=\frac{1x10^{-14}}{1x10^{-20}} =1x10^{-6}[/tex]
Next, the acid dissociation in the presence of the base we have:
[tex]Kb=\frac{[OH^-][HA]}{[A^-]}=1x10^{6}=\frac{x*x}{0.1-x}[/tex]
Whose solution is [tex]x=0.0999M[/tex] which equals the concentration of hydroxyl in the solution, thus we compute the pOH:
[tex]pOH=-log([OH^-])=-log(0.0999)=1[/tex]
Finally, since the maximum scale is 14, we can compute the pH by knowing the pOH:
[tex]pH+pOH=14\\\\pH=14-pOH=14-1\\\\pH=13[/tex]
Regards.
Answer:
pH = 13
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
