Question

1. (a) The human body contains4.7L of blood. If you drink 1 cup (237 mL) of orange juice, which contains 0.13M concentration of citric acid, what will be the pH of your blood? (You can consider your blood to be initially pure water.) The Kal of citric acid is 7.1 x 10-4 (b) Blood pH must remain between 7.35-7.45 for normal cellular metabolism. A pH < 6.8 or pH > 7.8 is not compatible with life. Comment on this and your answer from part (a). Can you come up with any explanations?

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Answer #1

Ans. #a. Total volume of body fluid after drinking orange juice = 4.7 L + 0.237 L

                                                = 4.937 L

# Using           C1V1 (original soln.) = C2V2 (diluted soln.)-

[Citric acid] in body fluid, C2 = (0.13 M x 0.237 L) / 4.937 L = 0.00624 M

# Calculate pH of body fluid using ICE approach as shown below-

Resultant pH of blood = 2.75

AH Weak acid Reaction: Given: Ka of AH = 7.100 E-04 Create an ICE table as shown in figure- A Conjugate base A + H3O 0 +X +X

#b. Change in blood pH beyond the range 7.35-7.45 causes denaturation of protein by interfering with the normal protein conformation due to addition or removal of H+ from charged amino acids in protein. Denaturation of protein further leads to loss of catabolic activities –thus shutting down most of the enzyme-catalyzed metabolic pathways in the organism. Shutting down of one or more metabolic pathways may lead to cell death, and subsequently death of the organism.

# Since lemon juice greatly lowers the pH of the blood (assuming blood acts as water, but no buffering activity is exhibited by blood in this case), consumption of large quantity of concentrated lemon juice may be fatal.

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