Human urine has a normal pH of 6.2. If a person eliminates an average of 1250.0

Question

Human urine has a normal pH of 6.2. If a person eliminates an average of 1250.0 mL of urine everyday, how man H^+ ions are eliminated in a week? Tris(hydroxymethyl) aminomethane, known as TRIS is a water-soluble base used in synthesizing surfactants and pharmaceuticals, as an emulsifying agent in cosmetics, and in cleaning mixtures for textiles and leather. In biomedical research, solutions of TRIS are used to maintain nearly constant pH for the study of enzymes and other cellular components. Given that the pK_b is 5.91, calculate the pH of 0.075 M TRIS.

Explanation / Answer

1) pH of Urine = 6.2

pH = -log[H+]

-log[H+] = 6.2

[H+] = 1×10^-6.2

= 6.3×10^-7M

= 6.3×10^-7mol/Litre

For 1250 litre = 6.3× 10^-7 × 1.25

= 7.88 × 10^-7 mol

For 7 days = 7 × 7.88 × 10^-7 mol

= 5.52 × 10^-6 mol

1 mol = 6.022 × 10^23 ions

No of H+ ions eliminated in a week=6.022×10^23×5.52× 10^-6

=3.32 × 10^17

2) Tris + H2O <-------> TrisH+ + OH-

pKb=5.91

Kb = 1.23 × 10^-6

Kb = [TrisH+][OH-]/[Tris]

1.23×10^-6 = X^2/(0.075-X)

We can assume (0.075 - X) = 0.075

X^2 = 9.22× 10^-8

X = 3.04×10^-4

[OH-] = 3.04 × 10^-4

pOH = 3.52

pH= 14 - pOH

= 14- 3.52

= 10.48

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