Question

PART A)

As a technician in a large pharmaceutical research firm, you need to produce 350. mL of 1.00 mol L−1 potassium phosphate buffer solution of pH = 6.93. The pKa of H2PO4− is 7.21.

You have the following supplies: 2.00 L of 1.00 mol L−1 KH2PO4 stock solution, 1.50 L of 1.00 mol L−1 K2HPO4 stock solution, and a carboy of pure distilled H2O.

How much 1.00 mol L−1 KH2PO4 will you need to make this solution?

Express your answer to three significant digits with the appropriate units.

PART B)

If the normal physiological concentration of HCO3− is 24 mmol L−1, what is the pH of blood if PCO2 drops to 31.0 mmHg ?

Express your answer numerically using two decimal places.

Answer 1

ans)

part a)

from baove data that

Let a be the volume of KH2PO4 and b be the volume of K2HPO4 needed in liters

Moles of KH2PO4 = volume x concentration = a x 1.00 = a mol

Moles of K2HPO4 = volume x concentration = b x 1.00 = b mol

we know that

Henderson-Hasselbalch equation:

pH = pKa + log([K2HPO4/[KH2PO4]])

= pKa + log(moles of K2HPO4/moles of KH2PO4) since final volume (= 400mL) is the same for both

6.93= 7.21 + log(b/a)

log(b/a) = -0.28

b/a = 10^-0.28

=0.524807

=> b = 0.524807 a

now we have to calculate the

Total moles of phophate = final volume x total concentration of phosphate

= 350/1000 x 1.00 =0.35mol

Thus a + b = 0.35

a + 0.5248a = 0.35

a = 0.2295

so finally the Volume of KH2PO4 needed = a =0.2295 = 229 mL

part b)