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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 300. mL of a potassium dihydrogen phosphate buffer solution of pH = 6.97. 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.)

User Sdfwer
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Answer:

The correct answer is 190.5 mL of 1.00 M KH₂PO₄

Step-by-step explanation:

A phosphate buffer is composed by phosphate acid (KH₂PO₄) and its conjugated base (K₂HPO₄). To obtain the relation between the concentrations of base and acid to add, we use Henderson-Hasselbach equation:

pH= pKa + log
(base)/(acid)

We have: pH= 6.97 and pKa= 7.21. So, we replace the values in the equation:

6.97= 7.21 + log
(base)/(acid)

6.97-7.21= log
(base)/(acid)

-0.24= log
(base)/(acid)


10^(-0.24)=
(base)/(acid)

0.575 =
(base)/(acid)


(0.575)/(1)=
(base)/(acid)

It means that you have to mix a volume 0.575 times of conjugated base and 1 volume of acid. If we assume a total buffer concentration of 1 M, we have:

base + acid = 1

base= 1 - acid

We replace in the previous equation:

0.575=
(1-acid)/(acid)

0.575 acid= 1 - acid

0.575 acid + 1 acid= 1

1.575 acid = 1

acid= 1/1,575

acid= 0.635

base= 1 - acid = 1 - 0.635 = 0.365

For a total volume of 300 ml, the volumes of both acid and base will be:

300 ml x 0.635 M = 190.5 ml of acid (KH₂PO₄)

300 ml x 0.365 M= 109.5 ml of base (K₂HPO₄)

We can corroborate our calculations as follows:

190.5 ml + 109.5 ml = 300 ml

109.5 ml / 190.5 ml = 0.575

User Ivan Kharlamov
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