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
the ph of the resulting solution when 1.15g of sodium metal is reacted slowly in 500cm3 of water is close to
2 answers
Sorry, used 1.50 gms Na instead of 1.15 gms Na. Just change the 1.5 to 1.15 and the problem is the same.
[Na⁰(s)]added = [NaOH]formed = [OH¯]ionized = 1.15g/23g-mol¯¹ = 0.05 M OH¯
pOH = -log[OH¯] = -log(0.05) = 1.3 => pH = 14 – pOH = 14 – 1.3 = 12.7
[Na⁰(s)]added = [NaOH]formed = [OH¯]ionized = 1.15g/23g-mol¯¹ = 0.05 M OH¯
pOH = -log[OH¯] = -log(0.05) = 1.3 => pH = 14 – pOH = 14 – 1.3 = 12.7