21.3 Weak acids and bases

Question 1

What are weak acids and bases

Answer 1

They only slightly Ionise when dissolved in water
Eg ethanoic acid

Question 2

In equation

CH3COOH <—> H+ + CH3COO-

explain how it dissociates

Answer 2

Most molecules remain bonded, only some dissociate into ions

So reaction stays at equilibrium because it doesn’t fully dissociate

Question 3

Explain equation

NH3 + H2O <—> NH4+ + OH-

Answer 3

Ammonia acts as a base and only partially dissociates so the NH3 can gain a proton, and water loses one

Question 4

What is equilibrium for a weak acid dissociating
If weak acid is HA
HA(aq) <—-> H+(aq) + A-(aq)

Answer 4

kc = [H+][A-] / [HA]

However weak acids are usually given symbol Ka and called the acid dissociation constant

ka = [H+(aq)][A-(aq)] / [HA(aq)]

Question 5

What does the size of Ka represent

Answer 5

The larger the value of ka, the further the equilibrium is to the right, because there are more products than reactants.

So the larger the value of ka, more of the acid is dissociated so the stronger it is.
Ka has units moldm^-3

Question 6

How do you calculate PH of weak acids

Answer 6

When calculating strong acids, you assume they are fully dissociated

However in weak acids you must use acid dissociation expression to calculate [H+]

Question 7

Calculate PH of 1moldm-3 ethanoic acid

CH3COOH <—> CH3COO-(aq) + H+(aq)

Value of ka for ethanoic is 1.7 x 10^-5

Answer 7

Ka = [CH3COO-][H+] / CH3COOH

But as each molecule of CH3COOH that dissociates produces one CH3COO- ion and one H+ ion:

[CH3COO-(aq)] = [H+(aq)]
Since the degree of dissociation of ethanoic acid is so small, the H+ concentration is very small so to a good approximation,
1-[H+] = 1

Ka= [H+(aq)]^2 / 1
so 1.7 x 10^-5 = [H+]^2
so [H+] = 4.12 x 10^-3 moldm-3

-log10(4.12 x 10^3) this equals 2.385 is PH

Question 8

What is pKa of a weak acid referred to as

Answer 8

pKa = -log10Ka
so p is -log10 of

So it is like pH, but for weak acid