Question

What will be the osmotic pressure of 0.02M monoacidic base if it's pH is 12 at 300 K?

Answer 1

First of all we have to find out concentration of [H⁺]
pH = -log[H⁺]
12 = -log[H⁺]
[H⁺] = 10⁻¹² M
∵[H⁺][OH⁻] = 10⁻¹⁴
[OH⁻] = 10⁻¹⁴/10⁻¹² = 10⁻² = 0.01M
Now, use [OH⁻] = Cα
Given, C = 0.02
so, α = 0.01/0.02 = 0.5
Now, use van Hoff's factor
i = α + 1
= 0.5 + 1 = 1.5
∴ osmotic pressure , π = iCRT
= 1.5 × 0.02 RT
= 0.03RT
Now, put R = 0.082 and T = 300K
π = 0.03 × 0.082 × 300
= 9 × 0.082 atm
= 0.738atm