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

Question

a. As a technician in a large pharmaceutical research firm, you need to produce 150. mL of 1.00 M potassium phosphate buffer solution of pH = 6.92. 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? Express your answer to three significant digits with the appropriate units.

b. If the normal physiological concentration of HCO3 is 24 mM, what is the pH of blood if PCO2 drops to 22.0 mmHg ?

Express your answer numerically using two decimal places.

Explanation / Answer

Buffer formula

pH = pKa + log [ Conjugate base] / [ Acid]

pH = pKa + log( HPO4-2] / [H2PO4-] )

log( [HPO4-2] / H2PO4-] ) = pH - pKa

[HPO4-2] / H2PO4-] = 10^(pH - pKa)

Cb x Vb / ( Ca x Va ) = 10^(pH - pKa)


Ca = Cb . . . therefore :

Vb / Va = 10^(pH - pKa)

Vb / Va = 10^(6.92 - 7.21)

Vb / Va = 0.512

. . . and :

Va + Vb = 0.15 L

Vb = 0.512 Va

Va + 0.512 Va = 0.15

Va = 0.0992 L = 99.2 ml

Vb = 0.0508 L =50.8 ml


. . . then, after solve :

Va = 99.2 ml of KH2PO4 . . . and . . . Vb = 50.8 ml of K2HPO4 is needed

b]

H2O + CO2 -----> H2CO3 ----> H+ + HCO3-

Modified form of buffer equation in blood

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

[HCO3-] in mM and PCO2 in mmHg

pH = 6.1 + log [24]*0.03*22 = 7.2997

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.