The Henderson-Hasselbalch equation relates the pH of a buffer solution to the p

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 200. mL of a potassium dihydrogen phosphate buffer solution of pH = 7.06. 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.)

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

The Henderson-Hasselbalch equation in medicine

Carbon dioxide (CO2) and bicarbonate (HCO3) concentrations in the bloodstream are physiologically controlled to keep blood pH constant at a normal value of 7.40.

Physicians use the following modified form of the Henderson-Hasselbalch equation to track changes in blood pH:

pH=pKa+log[HCO3](0.030)(PCO2)

where [HCO3] is given in millimoles/liter and the arterial blood partial pressure of CO2 is given in mmHg. The pKa of carbonic acid is 6.1. Hyperventilation causes a physiological state in which the concentration of CO2 in the bloodstream drops. The drop in the partial pressure of CO2 constricts arteries and reduces blood flow to the brain, causing dizziness or even fainting.

Part B

If the normal physiological concentration of HCO3 is 24 mM, what is the pH of blood if PCO2drops to 33.0 mmHg ?

Express your answer numerically using two decimal places.

Explanation / Answer

PART A:

Henderson–Hasselbalch equation
pH = pKa + log { [salt] / [acid] }
pH = pKa + log { [K2HPO4] / [KH2PO4] }
7.06 = 7.21 + log { [K2HPO4] / [KH2PO4] }
0.15 = log { [K2HPO4] / [KH2PO4] }
[K2HPO4] / [KH2PO4] = 100.15
[K2HPO4] / [KH2PO4] = 1.413

The ratio of concentrations has to be 1.413
But the ratio of concentrations in a solution, (200 mL), is the same as the ratio of moles in that volume, so I am going to convert the mLs of each stuff added into moles, where I arbitrarially choose one to be where that I added "x" millilitres of it , & the other becomes that I added "200 - x" mLs it by default:

1.413 = [K2HPO4] / [KH2PO4]

now I am going to multiply the molarity times the volume of each to get the moles of each , (& yes it won't matter that I amusing mls instead of liters):

1.413 = (200 mL - x) (1.00 M K2HPO4) / (x mL) (1.00 M KH2PO4)

1.413 = (200 - x) / (x)

1.413x = 200 - x

2.413x = 200 mL

x = 82.884 mL

So, the volumes are
K2HPO4 = 200 – x = 200 mL - 82.884 mL = 117.116 mL
KH2PO4 = x = 82.884 mL

So, answer is : mix 82.884 mL of 1.00 Molar KH2PO4 with 117.116 mL of 1.00 Molar K2HPO4

PART B

Using modified form of the Henderson-Hasselbalch equation

pH = pKa + log { [HCO3] (0.030) (PCO2) }
pH = 6.1 + log { [24 mM] (0.030) (33.0 mmHg) }
pH = 6.1 + log(23.76)
pH = 6.1 + 1.38
pH = 7.48   (upto two decimal places)

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