As a technician in a large pharmaceutical research firm, you need to produce 450

Question

As a technician in a large pharmaceutical research firm, you need to produce 450.mL of 1.00 M potassium dihydrogen phosphate buffer solution of pH = 6.89. 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?

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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.

If the normal physiological concentration of HCO3? is 24 mM, what is the pHof blood if PCO2 drops to 30.0mmHg ?

Explanation / Answer

The pH of a buffer can be calculated using Henderson equation which is given below.

pH = pKa + log ( Base / Acid )

Here, pH is given as 6.89

pKa is 7.21

Acid is KH2PO4 and base is K2PO4

After substituting the values, we get

6.89 = 7.21 + log ( K2HPO4 / KH2PO4)

-0.32 = log ( K2HPO4 / KH2PO4)

10^-0.32 = ( K2HPO4 / KH2PO4)

( K2HPO4 / KH2PO4) = 0.4786

In terms of moles , above equation can be expressed as

moles of K2HPO4 = 0.4786 * moles of KH2PO4 ............................... Equation 1

The volume required is 450 mL which is 0.45 L

Let's assume volume of KH2PO4 as x L

Therefore volume of K2HPO4 would be 0.45 - x L

Molarity = moles / L

Moles = Molarity * L

Molarity of both the solutions is 1 M

Therefore, moles of KH2PO4 = x mol

Moles of K2HPO4 = 0.45 - x

Let's plug these moles in equation 1

moles of K2HPO4 = 0.4786 * moles of KH2PO4

0.45 - x = 0.4786 * x

1.4786 x = 0.45

x = 0.304 L = 304 mL

304 mL of KH2PO4 is required to prepare the given buffer

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The equation to calculate buffer is already given

pH = pKa + log [ ( HCO3^- ) / 0.03 * pCO2 ] ............... ( Please check the equation posted. The correct equation should have log of ( HCO3^-) / 0.03 * pCO2)

pKa is given as 6.1

[HCO3^-] = 24 mM

pCO2 = 30 mm Hg

Let's plug in the values in modified Henderson equation

pH = 6.1 + log ( 24 / 0.03 * 30)

pH = 6.1 + 1.43

pH = 7.53

The pH of blood becomes 7.53 when pCO2 drops to 30.00 mmHg

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