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### How do buffers work?

What is a buffer?

A buffer is a solution that resists a change in pH when small amounts of acid or base are added. It is composed of either a weak acid or a weak base with its respective salt.

What's going on in the buffer?

For clarity, let's use the example ethanoic acid and sodium ethanoate.

CH3COOH (aq) ⇌ CH3COO- (aq) + H+ (aq) (1)

CH3COO-Na+ (aq) ⇌ CH3COO- (aq) + Na+ (aq) (2)

The acid and its salt partially and reversibly dissociate, creating the above equilibriums.

What is the concentration of the buffer?

[H+] = (Ka[CH3COOH]) / [CH3COO-Na+]

pH = - log([H+])

Note:

[acid] = mol. acid / V

[salt] = mol.salt / V

V = the total volume and so:

[acid] / [salt] = mol. acid / mol. salt

Therefore, [H+] = (Ka(mol. CH3COOH)) / (mol. CH3COO-Na+)

Adding acid

The H+ ions react with the CH3COO- ions.

CH3COOH (aq) ⇌ CH3COO- (aq) + H+ (aq) (1)

Equilibrium pushed , increasing [CH3COOH].

CH3COO-Na+ (aq) ⇌ CH3COO- (aq) + Na+ (aq) (2)

Equilibrium pushed , decreasing [CH3COO-Na+]

As all stoichiometric ratios are 1:1 and total volume is constant:

[H+] = (Ka(mol. CH3COOH + x)) / (mol. CH3COO-Na+ - x)

where x is the number of moles of acid added. As this is a small number, there is only a negligible change in pH = - log ([H+])

Adding a base

The alkali neutralises the acid.

CH3COOH (aq) ⇌ CH3COO- (aq) + H+ (aq) (1)

Equilibrium pushed , decreasing [CH3COOH].

CH3COO-Na+ (aq) ⇌ CH3COO- (aq) + Na+ (aq) (2)

Equilibrium pushed , increasing [CH3COO-Na+]

As all stoichiometric ratios are 1:1 and total volume is constant:

[H+] = (Ka(mol. CH3COOH - x)) / (mol. CH3COO-Na+ + x)

where x is the number of moles of acid added. As this is a small number, there is only a negligible change in pH = - log ([H+])

Note:

Ka = acid dissociation constant

[X] = concentration of X

mol. x = moles of x

and log = logarithm base 10

11 months ago

Answered by Daisy, an A Level Chemistry tutor with MyTutor

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