Question

The pKa of the phosphate buffer system ([H₂PO₄⁻]/[HPO₄⁻²]) is 6.8. What are the relative concentrations of [H₂PO₄⁻] and [HPO₄²⁻] in a urine sample that has a pH of 4.8.

Answer 1

Here first of all we have to identify which is acid and which one is base.
see reaction , H₂PO₄⁻ ⇔ HPO₄²⁻ + H⁺
so, acid = [H₂PO₄⁻]
base = [HPO₄²⁻]

Use the formula , pH = pKa + log₁₀ [base]/[acid]
Given, pH = 4.8 , pKa = 6.8
4.8 = 6.8 + log₁₀ [HPO₄²⁻]/[H₂PO₄⁻]
⇒ -2 = log₁₀ [HPO₄²⁻]/[H₂PO₄⁻]
⇒10⁻² = [HPO₄²⁻]/[H₂PO₄⁻]
So, [H₂PO₄⁻]/[HPO₄²⁻] = 1/0.01 = 100
Hence, relative concentration of [H₂PO₄⁻] and [HPO₄²⁻] = 100

Answer 2

Answer:

hey mate your answer is

Explanation:

100