Question

Enough of a monoprotic weak acid is dissolved in water to produce a 0.0122 M solution. The pH of the resulting solution is 2.64 . Calculate the Ka for the acid.

Answer 1

The pH of the resulting solution is 2.64.

Since pH is defined as the negative logarithm of [H₃O⁺], we can calculate the equilibrium H₃O⁺ concentration as follows:

Now, we can write the equation of an weak acid (HA) reacting with water as follows:

Now, given that the initial concentration of the weak acid HA is 0.0122 M, we can create the following ICE chart to calculate the equilibrium concentrations and thus the equilibrium constant K_a.

Initial, M	0.0122	0	0
Change, M	-x	+x	+x
Equilibrium, M	0.0122-x	x	x

Hence, for x moles of HA dissociating, we get x moles each of H₃O⁺ and A^-.

Note that water is the solvent and need not be included in the ICE chart and equilibrium expression.

Now, we already know that the equilibrium [H₃O⁺] is . Hence, using the ICE table, we can write

Hence,

Now, we can write the expression of equilibrium constant expression as follows:

Hence, the value of equilibrium constant K_a for the weak monoprotic acid is approximately .