Question

xFormula1" title="H_2PO_4^-(aq) \rightarrow H^+(aq) + HPO_4^{2-}(aq)" alt="H_2PO_4^-(aq) \rightarrow H^+(aq) + HPO_4^{2-}(aq)" align="absmiddle" class="latex-formula">
Find the pH of this buffer after 0.069 mol of OH- was added to 1.00 l of this buffer containing 0.165 M is $H_2PO_4^-$ and 0.594 $HPO_4^{2-}$ (Ka₁ = 7.2 x 10⁻³; Ka₂ = 6.1 x 10⁻⁸; Ka₃ = 4.2 x 10⁻¹³) assume no volume change due to the addition of OH⁻.

Answer 1

Answer:

The pH of the buffer solution = 8.05

Explanation:

Using the Henderson - Hasselbalch equation;

pH = pKa₂ + log ( [HPO₄²-]/[H₂PO4⁻]

where pKa₂ = -log (Ka₂) = -log ( 6.1 * 10⁻⁸) = 7.21

Concentration of OH⁻ added = 0.069 M (i.e. 0.069 mol/L)

[H₂PO4⁻] after addition of OH⁻ = 0.165 - 0.069 = 0.096 M

[HPO₄²-] after addition of OH⁻ = 0.594 + 0.069 = 0.663 M

Therefore,

pH = 7.21 + log (0.663 / 0.096)

pH = 7.21 + 0.84

pH = 8.05