Answer to Question #302374 in Chemistry for Gab

Question

The internal pH of a muscle is 6.8. Calculate the [H₂PO₄^-]/[HPO₄^2-] ratio in the cell. The second dissociation constant of phosphoric acid is 6.31x10^-8

Answer 1

Solution:

H₂PO₄⁻ ⇌ H⁺ + HPO₄²⁻, K₂ = 6.31×10⁻⁸

pK₂ = −log(K₂) = −log(6.31×10⁻⁸) = 7.1999 = 7.20

Henderson–Hasselbalch equation can be used.

pH = pKa + log([A⁻] / [HA])

pH = 6.80

pKa = pK₂ = 7.20

[A⁻] = HPO₄²⁻

[HA] = H₂PO₄⁻

Therefore,

6.80 = 7.20 + log([HPO₄²⁻] / [H₂PO₄⁻])

−0.40 = log([HPO₄²⁻] / [H₂PO₄⁻])

[HPO₄²⁻] / [H₂PO₄⁻] = 10^−0.40 = 0.398

[H₂PO₄⁻] / [HPO₄²⁻] = 1 / 0.398 = 2.51

[H₂PO₄⁻] / [HPO₄²⁻] = 2.51

Answer: The ratio [H₂PO₄⁻] / [HPO₄²⁻] is 2.51