Calculate the pH of 500cm3 of 0.2mol/dm3 Ca(OH2), assume complete ionisation

The pH of 500 cm³, 0.2 mol/dm³, Ca(OH₂) is ≈ 13.6

Explanation:

The given parameters are;

Concentration of the calcium hydroxide, Ca(OH₂) = 0.2 mol/dm³

Volume of the calcium hydroxide, Ca(OH₂) = 500 cm³

Given that the calcium hydroxide, Ca(OH₂),  undergoes complete ionization in water, complete dissociation, we have;

Ca(OH₂)  ⇄  Ca²⁺ + 2 OH⁻

Therefore, we have;

2 moles of [OH⁻] for each mole of Ca(OH₂) which gives the concentration of  [OH⁻] as

[OH⁻] = 2 × 0.2 ≈ 0.4 mol/dm³

The pH is given by the formula

pH = 14 + ㏒₁₀ [OH⁻]

pH = 14 + ㏒₁₀ (0.4) = 14 - 0.3979 = 13.6021 ≈ 13.6

The pH of the 500 cm³ 0.2 mol/dm³ Ca(OH₂) ≈ 13.6.

i'm assuming that the correct answer is indeed the first one/ "a"..