Question

The pH of the resulting solution when l.l5 g of sodium metal is reacted slowly in 500 cm3 of water is close to

Answer 1

Answer:

Rxn: Na⁰(s) + H₂O(l) => NaOH(aq) + ½H₂(g)

Moles Na⁰(s) = (1.50 gms / 23 gms/mol) = 0.065 mol Na⁰(s)

From Rxn ratios …

Moles Na⁰(s) consumed = Moles NaOH formed

=> 0.065 mol Na⁰(s) => 0.065 mol NaOH(aq)

=> [NaOH] = (0.065 mole NaOH / 0.500 Liter Solution) = 0.13M NaOH

=> Since NaOH is a strong Gp-IA Base, it will ionize 100% => 0.13M Na⁺(aq) + 0.13M OH¯(aq)

=> pOH = -log[OH¯(aq)] = -log(0.13) = 0.886

= > pH + pOH = 14 => pH = 14 – pOH = 14 – 0.886 = 13.11