Question

100 g of a 2.5 wt% solution of salt is concentrated to the solubility limit at 50°C. How much water remains in the concentrated liquid mixture for the following salts?

(a) Na₂SO₄

(b) Pb(NO₃)₂

(c) KCl

(d) Ce₂(SO₄)₃

Answer 1

(a) Solubility of Na2SO4 at 50 oC is 46.1 g Na2SO4/100 g water

2.5 wt% solution has = 2.5 g Na2SO4 in 97.5 g water

Saturated solution weight = 2.5 x 100/46.1 = 5.4 g solution

Amount of water present = 5.4 - 2.5 = 2.9 g

(b) Solubility of Pb(NO3)2 at 50 oC is about 15 g/100 g water

2.5 wt% solution has = 2.5 g Pb(NO3)2 in 97.5 g water

Saturated solution weight = 2.5 x 100/15 = 16.67 g

Amount of water present = 16.67 - 2.5 = 14.17 g

(c) Solubility of KCl at 50 oC 42.6 g KCl/100 g water

2.5 wt% solution has = 2.5 g KCl in 97.5 g

Saturated solution weight = 2.5 x 100/42.6 = 5.87 g

amount of water present = 5.87 - 2.5 = 3.37 g

(d) Solubility of Ce2(SO4)3 is about 3 g/100g water

saturated solution weight = 2.5 x 100/3 = 83.33 g

amount of water present = 83.33 - 2.5 = 80.83 g