# Sample problems salt solubility

16 Sample problems salt solubility

1. Problem: How many grams of NaCl (molar mass = 58.44 g/mol) can dissolve in 500 ml of water if the solubility is 357 g/L? Solution: Solubility = 357 g/L = 357 g/1000 ml = 0.357 g/ml, so in 500 ml it will be 0.357 g/ml x 500 ml = 178.5 g

2. Problem: How many moles of Na(2)SO(4) (molar mass = 142.04 g/mol) can dissolve in 1000 ml of water if the solubility is 18 g/100 ml? Solution: Solubility in 1L = 18 g/100 ml x 1000 ml = 180 g. Number of moles = 180 g / 142.04 g/mol = 1.27 moles
3. Problem: If the solubility product of AgCl is 1.77 x 10(-10) mol²/L², calculate the solubility in g/L (Molar mass of AgCl = 143.32 g/mol). Solution: [Ag⁺][Cl⁻] = Ksp, so solubility = sqrt(Ksp) = sqrt(1.77 x 10(-10)) = 1.33 x 10(-5) mol/L. So in g/L = 1.33 x 10(-5) mol/L x 143.32 g/mol = 1.91 x 10(-3) g/L
4. Problem: Calculate the solubility of CaCO(3) in g/L if Ksp = 4.8 x 10(-9) (Molar mass of CaCO(3) = 100.09 g/mol). Solution: [Ca²⁺][CO(3)²⁻] = Ksp, so solubility = sqrt(Ksp) = sqrt(4.8 x 10(-9)) = 6.93 x 10(-5) mol/L. So in g/L = 6.93 x 10(-5) mol/L x 100.09 g/mol = 0.0069 g/L
5. Problem: How many grams of PbI(2) (molar mass = 461.01 g/mol) can dissolve in 400 ml of water if the solubility is 0.06 g/100 ml? Solution: Solubility in 400 ml = 0.06 g/100 ml x 400 ml = 24 g
6. Problem: If 5 g of BaSO(4) (molar mass = 233.39 g/mol) is added to 100 ml of water, how many grams will remain undissolved if the solubility is 0.0024 g/100 ml? Solution: Soluble part = 0.0024 g/100 ml x 100 ml = 0.24 g. Undissolved part = 5 g – 0.24 g = 4.76 g
7. Problem: If the solubility of Al(OH)(3) in water is 0.00002 g/L, calculate the concentration in mol/L (Molar mass of Al(OH)(3) = 78.00 g/mol). Solution: 0.00002 g/L ÷ 78.00 g/mol = 2.56 x 10(-10) mol/L
8. Problem: If the solubility product of Mg(OH)(2) is 5.61 x 10(-12) mol²/L², calculate the solubility in g/L (Molar mass of Mg(OH)(2) = 58.32 g/mol). Solution: [Mg²⁺][OH⁻]² = Ksp, so solubility = (Ksp)^(1/3) = (5.61 x 10(-12))^(1/3) = 1.78 x 10(-4) mol/L. So in g/L = 1.78 x 10(-4) mol/L x 58.32 g/mol = 0.0104 g/L
9. Problem: How many moles of CuSO(4) (molar mass = 159.61 g/mol) can dissolve in 100 ml of water if the solubility is 203 g/L? Solution: Solubility in 100 ml = 203 g/L x 0.1 L = 20.3 g. Number of moles = 20.3 g / 159.61 g/mol = 0.127 moles
10. Problem: Calculate the molar solubility of PbCl(2) in water if Ksp = 1.7 x 10(-5) mol²/L². Solution: [Pb²⁺][Cl⁻]² = Ksp, so solubility = (Ksp)^(1/3) = (1.7 x 10(-5))^(1/3) = 0.00126 mol/L
11. Problem: Given the solubility of AgBr is 5.0 x 10(-7) mol/L, calculate the mass that can dissolve in 2 L of water (Molar mass of AgBr = 187.77 g/mol). Solution: Mass = solubility x volume x molar mass = 5.0 x 10(-7) mol/L x 2 L x 187.77 g/mol = 0.000188 g
12. Problem: Calculate the solubility of Fe(OH)(3) in mol/L given Ksp = 2.79 x 10(-39) (Molar mass of Fe(OH)(3) = 106.87 g/mol). Solution: Solubility = [Fe³⁺][OH⁻]³ = Ksp^(1/4) = (2.79 x 10(-39))^(1/4) = 3.22 x 10(-10) mol/L
13. Problem: Given 0.0015 mol of BaSO(4) (molar mass = 233.39 g/mol) can dissolve in 1 L of water, how many grams is this? Solution: Mass = moles x molar mass = 0.0015 mol x 233.39 g/mol = 0.35 g
14. Problem: How many grams of CuSO(4) (molar mass = 159.61 g/mol) can dissolve in 300 ml of water given the solubility is 900 g/L? Solution: Solubility = 900 g/L = 0.9 g/ml, so in 300 ml it will be 0.9 g/ml x 300 ml = 270 g
15. Problem: Calculate the solubility of ZnS in g/L given Ksp = 1.6 x 10(-24) (Molar mass of ZnS = 97.47 g/mol). Solution: Solubility = sqrt(Ksp) = sqrt(1.6 x 10(-24)) = 4.0 x 10(-13) mol/L. So in g/L = 4.0 x 10(-13) mol/L x 97.47 g/mol = 3.9 x 10(-11) g/L
16. Problem: Given the solubility product of AgI is 8.5 x 10(-17) mol²/L², calculate the solubility in g/L (Molar mass of AgI = 234.77 g/mol). Solution: Solubility = sqrt(Ksp) = sqrt(8.5 x 10(-17)) = 2.92 x 10(-9) mol/L. So in g/L = 2.92 x 10(-9) mol/L x 234.77 g/mol = 6.85 x 10(-7) g/L