Question

The phosphate buffer system is very important for maintaining the pH of the cytoplasm of all cells. Phosphoric acid is a triprotic acid; however, the relevant equilibrium in the biologically useful, neutral range, with a pKa of 6.86, is that of dihydrogen phosphate and monohydrogen phosphate ions: H3PO4 ⇌ H2PO4- + H+ pKa = 2.14 H2PO4- ⇌ HPO42- + H+ pKa = 6.86 HPO42- ⇌ PO43- + H+ pKa = 12.4 Using the Henderson-Hasselbalch equation, calculate the pH of a solution containing 0.042 M NaH2PO4 and 0.058 M Na2HPO4.

Answer 1

Answer:

The pH is 7.0

Explanation:

The equilibrium relevant for the problem is:

H₂PO4⁻ ⇌ HPO4⁻² + H⁺        pKa = 6.86

The Henderson-Hasselbalch (H-H) equation describes the pH of a buffer solution, using the concentrations of the acid and its conjugate base:

$pH=pka+log\frac{[A^{-} ]}{[HA]}$

For this case, [A⁻] = [HPO4⁻²], and [HA] = [H₂PO4⁻]. Thus:

$pH=pka+log\frac{[HPO4^{-2} ]}{[H2PO4^{-} ]}$

We put the concentrations and pka given in the problem in the H-H equation:

$pH=6.86+log\frac{0.058M}{0.042M}\\pH=6.86+log(1.381)\\pH=7.0$

P.S.: The actual pka for the equilibrium is 7.21, according to literature. Not 6.86 like the problem says. If we use a pka of 7.21 then the pH would be 7.35.