Answer:
Explanation:
The equilibrium relevant for this problem is:
H₂PO₄⁻ ↔ HPO₄⁻² + H⁺
The Henderson–Hasselbalch (H-H) equation is needed to solve this problem:
pH= pka + [tex]log\frac{[A^{-} ]}{[HA]}[/tex]
In this case, [A⁻] = [HPO₄⁻²], [HA] = [H₂PO₄⁻], pH = 7.4; from literature we know that pka=7.21.
We use the H-H equation to describe [HPO₄⁻²] in terms of [H₂PO₄⁻]:
[tex]7.4=7.21+log\frac{[HPO4^{-2} ]}{[H2PO4^{-} ]} \\0.19=log\frac{[HPO4^{-2} ]}{[H2PO4^{-} ]} \\10^{0.19}= \frac{[HPO4^{-2} ]}{[H2PO4^{-} ]} \\1.549*[H2PO4^{-} ]=[HPO4^{-2} ][/tex]
The problem tells us that the concentration of phosphate is 1 mM, which means:
[HPO₄⁻²] + [H₂PO₄⁻] = 1 mM = 0.001 M
In this equation we can replace [HPO₄⁻²] with the term expressed in the H-H eq:
1.549 * [H₂PO₄⁻] + [H₂PO₄⁻] = 0.001 M
2.549 * [H₂PO₄⁻] = 0.001 M
[H₂PO₄⁻] = 3.923 * 10⁻⁴ M
With the value of [H₂PO₄⁻] we can calculate [HPO₄⁻²]:
[HPO₄⁻²] + 3.923 * 10⁻⁴ M = 0.001 M
[HPO₄⁻²] = 6.077 * 10⁻⁴ M
With the concentrations, the molecular weight, and the volume, we calculate the mass of each reagent:
