Answer:
pH of a 0.10 M solution of OH⁻ = 12
pH of a 0.010 M solution of OH⁻ = 10
Explanation:
As the solution is basic, so the pH will be greater than 7
the pH of basic solution reduces by dilution
For 0.10 M solution of OH⁻
pOH = -log[OH⁻] = -log(0.10) = 1
pH = 14 – pOH = 14 – 1= 13.
For dilution by 10 factor, as the above value is for 1 L (so diluted 10 factor will be 100 ml)
c₁V₁=c₂V₂
c₂=c₁×V₁/V₂ = 0.10 mol/L × 100 mL/1000 mL = 10⁻³ mol/L
so now
pOH = -log[OH⁻] = -log (10⁻³) = 2.
pH = 14 – pOH = 14 – 2 = 12.
pH decreased from 13 to 12 for dilution by factor 10 of 0.10 Molar solution
For 0.010 M solution of OH⁻
pOH = -log[OH⁻] = -log(0.010) = 3
pH = 14 – pOH = 14 – 3= 11
For dilution by 10 factor, as the above value is for 1 L (so diluted 10 factor will be 100 ml)
c₁V₁=c₂V₂
c₂=c₁×V₁/V₂ = 0.010 mol/L × 100 mL/1000 mL = 10⁻⁴ mol/L
pOH = -log[OH⁻] = -log(10⁻⁴) = 4.
pH = 14 – pOH = 14 – 4 = 10
pH decreased from 13 to 10 for dilution by factor 10 of 0.010 Molar solution
