Question

Enough of a monoprotic acid is dissolved in water to produce a 0.0167 M solution. The pH of the resulting solution is 2.36. Calculate the Ka for the acid.

A certain weak base has a Kb of 7.50 × 10-7. What concentration of this base will produce a pH of 10.25?

Answer 1

First, use your pH to determine [H+] using 10^-pH= 10^-2.36=4.365x10^-3
Now, Ka=[H+][A-]/[HA]
However you can not disregard x in this case because it is close to your final concentration.
So, you have (4.365x10^-3)^2/0.0167-(4.365x10^-3)
Ka=1.545x10^-3

Kb= 7.50 × 10-7

The equilibrium reaction is below
B + H2O <=> CA + HO-
Equilibrium equation:
Kb= [HO-][CA]/[B]
[B]= concentration of base [CA]= concentration of conjugate acid
Also [HO-]= [CA] = x

14 - pH = pOH............................14-10.25 = 3.75
-log[HO-] = pOH................................... -log[HO-]= 3.75
[HO-] = 10^ (-pOH) .............................. [HO-] = 10^-3.75 = 1.778 x 10^-4

Since [HO-]= [CA] = x and we now know that x is 1.778 x 10^-4

So using the equilibrium equation and replacing x for [HO-] and [CA] we get:

Kb = x^2/[B]

Solve for [B]

[B] = x^2/Kb

[B] = (1.778 x 10^-4)^2 / (7.5 × 10-7)

[B]= 0.0422