Question

Calculate the pH of a mixture containing 100 ml of .2M KH2PO4 and 150 ml of 0.1 M K2HPO4.

Answer 1

KH2PO4 + K2HPO4 will form a buffer;

let H2PO4- be the weak acid HA and HPO4-2 the conjugate base A-

knowing this:

This is an acidic buffer; since there is a weak acid + conjugate base:

A buffer is any type of substance that will resist pH change when H+ or OH- is added.

This is typically achieved with equilibrium equations. Both type of buffer will resist both type of additions.

When a weak acid and its conjugate base are added, they will form a buffer

The equations that explain this phenomena are given below:

The Weak acid equilibrium:

HA(aq) <-> H+(aq) + A-(aq)

Weak acid = HA(aq)

Conjugate base = A-(aq)

Neutralization of H+ ions:

A-(aq) + H+(aq) <-> HA(aq); in this case, HA is formed, H+ is neutralized as well as A-, the conjugate

Neutralization of OH- ions:

HA(aq) + OH-(aq) <-> H2O(l) + A-(aq) ; in this case; A- is formed, OH- is neutralized as well as HA.

Note that the equilibirum equation can be mathematically manipulated in order to favour the "buffer" construction

Recall that, in equilibrium

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

Multiply both sides by [HA]

[HA]*Ka = [H+][A-]

take the log(X)

log([HA]*Ka) = log([H+][A-])

log can be separated

log([HA]) + log(Ka) = log([H+]) + log([A-])

note that if we use "pKx" we can get:

pKa = -log(Ka) and pH = -log([H+])

substitute

log([HA]) + log(Ka) = log([H+]) + log([A-])

log([HA]) + -pKa = -pH + log([A-])

manipulate:

pH = pKa + log([A-]) - log([HA])

join logs:

pH = pKa + log([A-]/[HA])

or better:

pH = pKa2 + log(HPO4-2/H2PO4-)

pKa2 = 7.21 for the 2nd ionization (HPO4-2/H2PO4-)

get mmol of each sample:

mmol of KH2PO4 = M*V = 0.2*100 = 20

mmol of K2HPO4 = MV = 150*0.1 = 15

now, substitute

pH = pKa2 + log(HPO4-2/H2PO4-)

pH = 7.21 + log(15/20)

pH = 7.085061