Question

part A :

As a technician in a large pharmaceutical research firm, you need to produce 400. mL of a potassium dihydrogen phosphate buffer solution of pH = 6.76. The pKa of H2PO4− is 7.21.

You have the following supplies: 2.00 L of 1.00 M KH2PO4 stock solution, 1.50 L of 1.00 M K2HPO4 stock solution, and a carboy of pure distilled H2O.

How much 1.00 M KH2PO4 will you need to make this solution? (Assume additive volumes.)

part B:

If the normal physiological concentration of HCO3− is 24 mM, what is the pH of blood if PCO2drops to 24.0 mmHg ?

Answer 1

let

y L of K2HP04 be taken

then

volume of KH2P04 = 0.40 - y

we know that

moles = molarity x volume (L)

so

moles of K2HP04 = 1 x y = y

moles of KH2P04 = 1 x (0.40-y) = 0.35-y

now

we know that

for buffers

pH = pKa + log [base /acid]

so

pH = pKa + log [ K2HP04 / KH2P04]

6.76 = 7.21 + log [ K2HP04 / KH2P04]

[ K2HP04 / KH2P04] = 0.3548

as the final volume is same for both

moles of K2HP04 / KH2P04 = 0.3548

moles of K2HP04 = 0.60256 x moles of KH2P04

so

y = 0.3548 x ( 0.40 - y)

y = 0.10476

so

volume of K2HP04 = 0.10476 L = 104.76 ml

volume of KH2P04 = 400 - 104.76 = 295.24 ml

B)

given

pH = pKa + log [HC03-] / (0.03 x PC02)

given

[HC03-] = 24 mM

pC02 = 24

pKa = 6.1

so

using those values

we get

pH = 6.1 + log [ 24 / 0.03 x 24]

pH = 7.623

so

pH of the blood is 7.623