From solubility product grams NaCl calculate (Salze-76) The solubility product of NaCl of which: Lp(NaCl) = 18.26 mol2/l2. Place two saturated fresh or chilled-solutions with the following: volumes I 'm here: a) 1.0 litres, b 0.3 litres

  Exercise: Salze-76    

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The solubility product of NaCl of which: L_p(NaCl) = 18.26 mol²/l². Place two saturated fresh or chilled-solutions with the following: volumes I 'm here: a) 1.0 litres, b 0.3 litres

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Solution:

It shall apply:
— 1·NaCl ⇄ 1·Na⁺ + 1·Cl^–
— c(NaCl) = c(Na⁺) = c(Cl^–)
— L_p(NaCl) = 18.26 mol²/l² = c(Na⁺)·c(Cl^–) = c²(Na⁺) = c²(Cl^–) = c²(NaCl)
–> c(NaCl) = √(L_p) = 4.27 mol/l
From c=n/V= m/M/V      following:: m = c·V·M
for 1.0 litres: m(NaCl) = c(NaCl) ·V·M(NaCl) = 4.27 mol/l·1.0 l · 58.5 g/mol = 249.8 g
for 0.3 litres: m(NaCl) = c(NaCl) ·V·M(NaCl) = 4.27 mol/l·0.3 l · 58.5 g/mol = 74.9 g

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