Question

pka is 7.21, Assigned pH: ____6.8__________ . prepare 50 mL of a 50 mM buffer at your assigned pH, calculate the ratios of the monobasic and dibasic species you will be using with the Henderson-Hasselbalch equation. Ratios: Monobasic species: ______________ Dibasic species: ______________ Volumes: Monobasic species: ______________ Dibasic species: ______________

Answer 1

For apKa = 7.21

this is most likely H3PO4, since it has its pKa2 = 7.21

If pH goal s 6.80 then

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])

substitute all data

in this case, A- is actually the acid HPO4-2 and HA is the acid H2PO4-

6.80 = 7.21 + log([HPO4-2]/[H2PO4-])

total V = 50 mL, total M = 50 mM = 50*10^-3 = 0.05 M

total mmol = MV = 50*0.05 = 2.5 mmol of buffer

EQN(1) =

mmol of HPO4-2 + mmol of H2PO4- = 2.5 mmol

From EQN(2) :

6.80 = 7.21 + log([HPO4-2]/[H2PO4-])

10^(6.80-7.21) = [HPO4-2]/[H2PO4-]

0.3890= [HPO4-2]/[H2PO4-]

since they are n the same volume, the ratio is the same

let

mmol of HPO4-2 = x and mmol of H2PO4- = y

mmol of HPO4-2 + mmol of H2PO4- = 2.5 mmol

x + y = 2.5

0.3890= x/y

then

x = 0.3890*y

substitute:

x + y = 2.5

0.3890*y + y = 2.5

y = 2.5/(1.389) = 1.79985

then

x + y = 2.5

x + 1.79985 = 2.5

x = 2.5-1.79985 = 0.70

that implies

in a V = 50 mL solution add:

use "sodium salts" since those are the most typical ones

MW of Na2HPO4 = 141.9588

mass of Na2HPO4 = mmol*MW = 0.70*141.9588 = 99.371 mg of Na2HPO4 must be added

MW of NaH2PO4 = 119.9770

mass of NaH2PO4 = mmol*MW = 1.799*119.9770 = 215.8 mg of NaH2PO4 must be added

note that the pH will be pretty near to 6.80