Question

The equilibrium constant expression for a weak acid is:

Ka=[HX_3OX^+]eq[AX^−]eq/[AH]eq for the reaction AH(aq) + H_2O(l)⇌HX3_OX^+(
)+AX^− If you solve for the hydronium ion activity and take the logarithm, you get: pH = pKa + log[AX−]initial[AH]initial[AX −] initial [AH] initial
​
a) True
b) False

Answer 1

The statement in the question is false. The correct equation is pH = pKa + log([AX−]initial/[AH]initial).

Step-by-step explanation:

The statement in the question is false.

The equilibrium constant expression for a weak acid is given by Ka = [HX3OX+]eq[AX−]eq/[AH]eq.

If we take the logarithm of this expression, we get: pH = pKa + log([AX−]initial/[AH]initial).

So, the correct equation is pH = pKa + log([AX−]initial/[AH]initial).