Unsure on how to work out this acid and base question?

pH = -log[H⁺]

Then, [H⁺] = 10^(-pH) M


Before dilution: C₁ = 10⁻⁰˙⁵ M (i.e. pH = 0.5), V₁ = 25 cm³

After dilution: C₂ = 10⁻⁰˙⁷ M (i.e. [H = 0.7], V₂ = ? cm³


For dilution: C₁V₁ = C₂V₂

Then, V₂ = V₁ × (C₁/C₂)


Final volume, V₂ = (25 cm³) × (10⁻⁰˙⁵/10⁻⁰˙⁷) = 39.6 cm³


Assume that the volume of the solution is additive.

Volume of water added = (39.6 - 25) cm³ = 14.6 cm³


[The given answer (12.5 cm³) is incorrect.]

Dilution ...

pH = 0.500 .... [H+] = 10^-0.500 = 0.316M
pH = 0.700 .... [H+] = 0.200M

0.025L x (0.316 mol H+ / 1L) x (1L / 0.200 mol H+) = 0.0395L

0.0395L - 0.0250L = 0.0145L ... or .... 14.5 mL

14.5 mL is the volume increase needed to drop the pH from 0.5 to 0.7.

Calculate H^+ at pH of 0.5 and pH of 0.7

pH = 10^-0.5 = 0.316228 M <--- carry some extra digits
pH = 10^-0.7 = 0.199526 M

Use M1V1 = M2V2 to calculate the dilution

(0.316228 mol/L) (25 cm^3) = (0.199526 mol/L) (x)

x = 39.6 cm^3

Subtract

39.6 - 25 = 14.6 cm^3 added

Not sure how to obtain 12.5 for the answer.