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Mekhanik [1.2K]
3 years ago
5

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

Chemistry
2 answers:
Vedmedyk [2.9K]3 years ago
5 0

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.

Agata [3.3K]3 years ago
3 0

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

<em>Where X is the reaction coordinate</em>

Replacing in Kb formula:

1x10⁶ = [X] [X] / [0.1-X]

1x10⁵ - 1x10⁶X = X²

<em>0 = X² + 1x10⁶X - 1x10⁵</em>

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

<em>pH = 13</em>

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