Part A As a technician in a large pharmaceutical research firm, you need to prod

Question

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 = 6.88. 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 34.0 mmHg ?

Explanation / Answer

A)

V = 350 mL of KH2PO4 buffer, pH = 6.88

pH is nearest to pKa2 = 7.21, i.e. we need H2PO4- and HPO4-2

so.. let (1) be the KH2PO4, and (2) be the HPO4-

Vtotal = V1+V2

pH = pKa2 + log(HPO4-2 / H2PO4-)

6.88 = 7.21 + log(HPO4-2 / H2PO4-)

(HPO4-2 / H2PO4-) = 10^(6.88-7.21) = 0.4677351

mmol of HPO4-2 = M2*V2 = 1*V2

mmol of H2PO4- = M1*V1 = 1*V1

350 = V1+V2

1*V2 / ( 1*V1) = 0.4677351

V2 = 0.4677351*V1

substitute

350 = V1+V2

350 = V1+0.4677351*V1

V1 = (350)/(1.4677351) = 238.46 mL

V2 = 350-238.46= 111.54 mL

So.. we need 238.5 mL of the KH2PO4 solution, and 111.5 mL of K2HPO4 solution

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