2 A 100 L solution is prepared using 00167 mol of K2HPO4 and

2. A 1.00 L solution is prepared using 0.0167 mol of K2HPO4 and 0.05 mol of NaH2PO4. a. Calculate the pH of this solution. Tot (nuvotin t00wan .1S ov .IS b. Recalculate the pH using activities (including the final calculation of the pH)

Solution

2.

a)

K₂HPO₄ --> 2K⁺ + HPO₄^2-

NaH₂PO₄ --> Na⁺ + H₂PO₄^-

[HPO₄^2-] = 0.0167 M

[H₂PO₄^-] = 0.05 M

[K⁺] = 2 * 0.0167 M = 0.0334 M

[Na⁺] = 0.05 M

pKa of H₂PO₄^- = 7.2

pH = pKa + log([HPO₄^2-]/[H₂PO₄^-])

= 7.2 + log(0.0167/0.05)

= 6.7

[H⁺] = 10^-6.7 = 1.89 x 10^-7

b)

Ionic strength, µ = ½ (CiZi²)

= ½ * ([HPO₄^2-] * 4 + [H₂PO₄^-] * 1 + [K⁺] * 1 + [Na⁺] * 1)

= 0.1

Using the Extended Debye-Huckel we get,

log (_H+) = -0.51 * 1 * µ^1/2 / (1 + µ^1/2)

= -0.51 * 0.1^1/2 / (1 + 0.1^1/2)

= -0.12

_H+ = 0.75

Activity of H⁺, aH⁺ = _H+ * [H⁺] = 0.75 * 1.89 x 10^-7

= 1.42 x 10^-7

pH = -log (aH⁺) = 6.85