Question

The Henderson-Hasselbalch equation relates the pH of a buffer solution to the pKa of its conjugate acid and the ratio of the concentrations of the conjugate base and acid. The equation is important in laboratory work that makes use of buffered solutions, in industrial processes where pH needs to be controlled, and in medicine, where understanding the Henderson-Hasselbalch equation is critical for the control of blood pH.

Part A As a technician in a large pharmaceutical research firm, you need to produce 350. mL of a potassium dihydrogen phosphate buffer solution of pH = 6.80. 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 PCO2 drops to 35.0 mmHg ? Express your answer numerically using two decimal places.

Answer 1

1) Molarity of KH₂PO₄ stock solution = 1 M and its volume is 2 L

Molarity of K₂HPO₄ stock solution = 1M and its volume is 1.5 L

Concentration of HPO₄^2- ,(K₂HPO₄), anion required be A^-

Concentration of H₂PO₄^- (KH₂PO₄) , acid required HA

According to Henderson -Hasselbalch equation :

pH = pKa + log [A]/[HA]

6.8 = 7.2 + log [A]/[HA]

6.8-7.2 = log [A]/[HA]

-0.4 = log [A]/[HA]

[A]/[HA] = 10^-(0.4) = 0.398

Total volume of buffer = 350 ml

Let us consider the volume of acid used by X

So Volume of HPO₄² will be 350-X

[A]/[HA] = 0.398 = (350-X) x 1 M / X x 1 M

0.398 X = 350-X

0.398 X+X = 350

1.398 X = 350

X = 350/1.398 = 250.36 ml of 1 M KH₂PO₄ acid from 2 L is used

350-X = 99.64 ml of 1 M K₂HPO₄ from 1.5 L is used

or it can be estimated as following :

[A]/[HA] = 0.398

Moles of [A]+[HA] = 350 ml x 1M = 350 mmoles

[A] = 0.398 [HA]

1.398 M [HA] = 350 mmoles

[HA] = 350 mmoles/1.398 M = 250.36 ml of 1 M KH₂PO₄ acid from 2 L is used

[A^-] = 0.398 X 250.36 ml = 99.64 ml of 1 M K₂HPO₄ acid from 1.5 L is used

B) concentration of HCO₃⁻ is 24 mM

pCO2 drops to 35.0 mmHg

Henderson-Hasselbalch equation of blood pH :

pH = pKa + log [A^-]/[HA]

pH = 6.1 + log [HCO₃^-]/0.03 x pCO₂

pH = 6.1 + log 24/(0.03 x 35)

pH = 6.1 + log 24/1.05

pH = 6.1 + log 22.86

pH = 6.1 + 1.36 = 7.46