A certain acid,HA,has a K_a given by:
HA + H₂O ⇌ H₃O⁺ + A^- K_a = 6.80 × 10^-6
What is the pH of a 0.247 M aqueous solution of the acid's potassium salt,KA,which undergoes the hydrolysis reaction?
A^- + H₂O ⇌ OH^- + HA

Question

Quizplus · Accepted Answer

First, we need to write the equation for the hydrolysis of the potassium salt:
KA + H2O ⇌ KOH + HA
Next, we need to write the expression for the equilibrium constant, K:
K = [KOH][HA]/[KA]
We know that KA = 0.247 M, and we can assume that KOH is a strong base and will be fully dissociated. Therefore, [KOH] = [OH-] = x, and [HA] = (6.02 × 10^-7)/x. (6.02 × 10^-7 is the concentration of HA calculated using the given equilibrium constant.)

Substituting these values into the expression for K, we get:
K = x(6.02 × 10^-7)/0.247

Solving for x, we get:
x = 1.72 × 10^-6 M

Finally, we can use this value of [OH-] to calculate the pH:
pH = 14 - pOH = 14 - (-log[OH-]) = 14 - (-log[1.72 × 10^-6]) = 9.28

Therefore, the correct answer is B.

A Certain Acid,HA,has a Ka Given By

A Certain Acid,HA,has a K_a Given By