Question

Consider a basic solution diluted by a factor of 10. Calculate the pH of a

0.10 M solution of OH⁻ and a 0.010 M solution of OH⁻. How did the pH
change? Please help immediately this is another question that is part of my lab!!!:(

Answer 1

pH of a 0.10 M solution of OH⁻ = 12

pH of a 0.010 M solution of OH⁻ = 10

Step-by-step 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