Question

At 25 ∘C phosphoric acid, H3PO4, has the following equilibrium constants:
H3PO4(aq)+H2O(l)H2PO4−(aq)+H2O(l)HPO42−(aq)+H2O(l)⇌⇌⇌H3O+(aq)+H2PO−4(aq)H3O+(aq)+HPO42−(aq)H3O+(aq)+PO43−(aq) Ka1Ka2Ka3===7.5×10−36.2×10−84.2×10−13

What is the pH of a solution of 0.900 M KH2PO4, potassium dihydrogen phosphate?

Answer 1

Question #79162

At 25 °C phosphoric acid, H3PO4, has the following equilibrium constants: H3PO4(aq)+H2O(l)H2PO4-(aq)+H2O(l)HPO42-(aq)+H2O(l)⇌⇌H3O+(aq)+H2PO-4(aq)H3O+(aq)+HPO42-(aq)H3O+(aq)+PO43-(aq) Ka1Ka2Ka3===7.5×10-36.2×10-84.2×10-13

What is the pH of a solution of 0.900 M KH2PO4, potassium dihydrogen phosphate?

The right answer is 4.67.

Solution:

Potassium dihydrogen phosphate is an ampholyte [1], so the formula of pH for it is equal to [2]:

$$pH = \frac{pK_{a,H_3PO_4} + pK_{a,H_2PO_4}^-}{2} = \frac{-\log(7.5 * 10^{-3}) - \log(6.2 * 10^{-8})}{2} = 4.67$$

So, the right answer is 4.67.

References:

[1] https://en.wikipedia.org/wiki/Amphoterism

[2] http://theochem.ki.ku.dk/~axhun/pHnoter.pdf