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.

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the seeing the electronic configuration we can say that the element is fluorine.

fluorine is the most electronegative element in the periodic table.

but by only seeing the configuration we cannot say about its reactivity because we don't know if its configuration after excitation or by addition of electrons.

but if not then, it is the lightest halogen and exists as a highly toxic pale yellow diatomic gas at standard conditions. as the most electronegative element, it is extremely reactive: almost all other elements, including some noble gases, form compounds with fluorine.