(a) The molar solubility of Ag2CrO4(s) at 35°C is 1.3 10-4 mol/L. Calculate Ksp.
(b) If 0.0490 g of AgIO3 dissolves per liter of solution, calculate the solubility-product constant.
(c) Using the appropriate Ksp value from Appendix D, calculate the solubility of Cu(OH)2 in grams per liter of solution.
3 g/L
oops...
(c) Using the appropriate Ksp value from Appendix D, calculate the solubility of Cu(OH)2 in grams per liter of solution.
Appendix D: Ksp for Cu(OH)2 = 4.8 x 10-20
I got A I just need b and c:)
I'm curious. If you did A, then B and C should be a snap. What's the problem?
To calculate the solubility-product constant (Ksp) or the solubility of a compound, we need to use the balanced equation for the dissolution reaction and the given molar solubility or mass of the compound that dissolves.
(a) To calculate the Ksp for Ag2CrO4(s), we need to write the balanced equation for its dissolution in water:
Ag2CrO4(s) ⇌ 2 Ag+(aq) + CrO4^2-(aq)
From the balanced equation, we can see that the stoichiometric coefficients for Ag2CrO4, Ag+, and CrO4^2- are 1, 2, and 1, respectively.
Given that the molar solubility of Ag2CrO4(s) is 1.3 x 10^-4 mol/L, we can plug this value into the equilibrium expression for Ksp:
Ksp = [Ag+]^2 [CrO4^2-]
Since the stoichiometric coefficients are 1:2, the concentration of Ag+ is twice that of CrO4^2-. Therefore:
Ksp = (2x)^2 (x) = 4x^3
Substituting the molar solubility, we get:
Ksp = 4(1.3 x 10^-4)^3 ≈ 1.27 x 10^-11
Thus, the Ksp for Ag2CrO4(s) at 35°C is approximately 1.27 x 10^-11.
(b) To calculate the solubility-product constant (Ksp) for AgIO3, we need to find the molar solubility using the mass given.
Given that 0.0490 g of AgIO3 dissolves per liter of solution, we calculate the molar solubility:
Molar solubility = mass of AgIO3 / molar mass of AgIO3
The molar mass of AgIO3 can be calculated by adding the atomic masses of Ag, I, and O:
Ag: 107.87 g/mol
I: 126.90 g/mol
O: 16.00 g/mol (three oxygens)
Molar mass of AgIO3 = 107.87 + 126.90 + (16.00 x 3) = 234.77 g/mol
Substituting the values:
Molar solubility = 0.0490 g / 234.77 g/mol ≈ 2.09 x 10^-4 mol/L
Now, we can calculate the Ksp using the molar solubility:
Ksp = [Ag+][IO3^-]
Since the stoichiometric coefficients are 1:1, we can write:
Ksp = (2.09 x 10^-4)^2 = 4.36 x 10^-8
Therefore, the solubility-product constant (Ksp) for AgIO3 is approximately 4.36 x 10^-8.
(c) To calculate the solubility of Cu(OH)2 in grams per liter of solution, we need to use the given Ksp value from Appendix D and the balanced equation for its dissolution:
Cu(OH)2(s) ⇌ Cu2+(aq) + 2 OH^-(aq)
The stoichiometric coefficients for Cu(OH)2, Cu2+ and OH- are 1, 1, and 2, respectively.
Given that Ksp for Cu(OH)2 is 2.2 x 10^-20, we can use the equilibrium expression to find the concentration of Cu2+ and OH-:
Ksp = [Cu2+][OH^-]^2
Since the stoichiometric coefficients are 1:2, the concentration of OH- is twice that of Cu2+. Therefore:
Ksp = (x)(2x)^2 = 4x^3
From the Ksp value, we have:
2.2 x 10^-20 = 4(2x)^3
Simplifying the equation, we find:
2.2 x 10^-20 = 32x^3
Taking the cube root of both sides, we get:
x = (2.2 x 10^-20 / 32)^(1/3) ≈ 1.97 x 10^-7 mol/L
Now, we convert the molar solubility to grams per liter using the molar mass of Cu(OH)2:
Molar mass of Cu(OH)2 = (63.55 g/mol) + 2(16.00 g/mol) + 2(1.01 g/mol) = 97.55 g/mol
Solubility = (1.97 x 10^-7 mol/L) * (97.55 g/mol) ≈ 1.92 x 10^-5 g/L
Therefore, the solubility of Cu(OH)2 in grams per liter of solution is approximately 1.92 x 10^-5 g/L.