Question

Consider an exceptionally weak acid, HA, with a Ka = 1x10 -20 . You make a 0.1M solution of the salt Na

Answer 1

Answer:

$pH=13$

Explanation:

Hello,

In this case, given the acid, we can suppose a simple dissociation as:

$HA\rightleftharpoons H^+ + A^-$

Which occurs in aqueous phase, therefore, the law of mass action is written by:

$Ka=\frac{[H^+][A^-]}{[HA]}$

That in terms of the change $x$ due to the reaction's extent we can write:

$1x10^{-20}=\frac{x*x}{0.1M-x}$

But we prefer to compute the Kb due to its exceptional weakness:

$Kb=\frac{Kw}{Ka}=\frac{1x10^{-14}}{1x10^{-20}} =1x10^{-6}$

Next, the acid dissociation in the presence of the base we have:

$Kb=\frac{[OH^-][HA]}{[A^-]}=1x10^{6}=\frac{x*x}{0.1-x}$

Whose solution is $x=0.0999M$ which equals the concentration of hydroxyl in the solution, thus we compute the pOH:

$pOH=-log([OH^-])=-log(0.0999)=1$

Finally, since the maximum scale is 14, we can compute the pH by knowing the pOH:

$pH+pOH=14\\\\pH=14-pOH=14-1\\\\pH=13$

Regards.

Answer 2

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