Question

Your TA needs to make 250 ml of 0.15 M phosphate buffer (pH 7.2), but the pH meter is not available. How much of each chemical (acid and conjugate base) should you weigh out to come close? MW of KH2PO4=136 K2HPO4=174

Answer 1

Vtotal = 250 mL

total molarity = 0.15 M

then, total moles of buffer = MV = 0.15*250 = 37.5 mmol of phosphate

Now; recall that for a buffer:

The Buffer Equation (Henderson Hasselbach) ACIDIC

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])

which is Henderson hasselbach equations.

For KH2PO4; K2HPO4

we have H2PO4- and HPO4-2 acids, which is the 2nd ionization of the H3PO4 acid

pKa2 = 7.21

then;

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

7.20 = 7.21 + log(HPO4-2 / H2PO4-)

10^(7.20-7.21) = (HPO4-2 / H2PO4-)

0.978 = (HPO4-2 / H2PO4-)

0.978 *H2PO4- = HPO4-2

and we know that

HPO4-2 + H2PO4- = 37.5

then, solve:

0.978 *H2PO4- = HPO4-2

0.978 *H2PO4- + H2PO4- = 37.5

(1+0.978 ) * H2PO4- = 37.5

H2PO4- = 37.5/(1+0.978 ) = 18.9585 mmol

HPO4-2 + H2PO4- = 37.5

HPO4-2 = 37.5 - 18.9585 = 18.5415 mmol

Now,

K2HPO4 = 18.5415 mmol

KH2PO4 = 18.9585 mmol

change to mass

mass of K2HPO4 = mmol*Mw = (18.5415 *174) = 3226.221 mg = 3.223 grams

mass of KH2PO4 = mmol*Mw = (18.9585 *136) = 2578.356 mg = 2.578 grams