Calculate the molar solubility of Ag2SO4 when dissolved in water (Ksp = 1.8 x 10-5)
0.0235
To calculate the molar solubility of Ag2SO4 when dissolved in water, we need to use the solubility product constant (Ksp) and set up an equilibrium expression.
The balanced equation for the dissociation of Ag2SO4 is:
Ag2SO4(s) ↔ 2Ag+(aq) + SO4^2-(aq)
The Ksp expression for this equilibrium is:
Ksp = [Ag+]^2[SO4^2-]
Given that the value of Ksp is 1.8 x 10^(-5), we can assume that the concentration of Ag+ and SO4^2- at equilibrium is "x". Thus, we can write the Ksp expression as:
1.8 x 10^-5 = (2x)^2(x)
Simplifying the equation:
1.8 x 10^-5 = 4x^3
Next, we solve for "x":
x^3 = (1.8 x 10^-5)/4
x^3 = 4.5 x 10^-6
Taking the cube root of both sides:
x = (4.5 x 10^-6)^(1/3)
x ≈ 0.0154 mol/L
Therefore, the molar solubility of Ag2SO4 when dissolved in water is approximately 0.0154 mol/L.
To calculate the molar solubility of a compound when dissolved in water, we need to use the concept of solubility product constant (Ksp).
The chemical equation for the dissociation of Ag2SO4 in water is:
Ag2SO4(s) ↔ 2 Ag+(aq) + SO4^2-(aq)
According to the equation, for every 1 mole of Ag2SO4 that dissolves in water, 2 moles of Ag+ ions and 1 mole of SO4^2- ions are formed. Therefore, the molar solubility of Ag2SO4(s) can be represented by "x" in moles per liter (M).
Ksp represents the product of the concentrations of the ions in a saturated solution. Since the stoichiometry of the compound indicates that 1 mole of Ag2SO4 produces 2 moles of Ag+ ions, the concentration of the Ag+ ions is expressed as 2x. Similarly, the concentration of SO4^2- ions is expressed as x.
Hence, the Ksp expression for Ag2SO4 is:
Ksp = [Ag+]^2 x [SO4^2-]
Substituting the expressions for [Ag+] and [SO4^2-], we get:
Ksp = (2x)^2 x x = 4x^3
Given that Ksp = 1.8 x 10^-5, we can now solve for x:
1.8 x 10^-5 = 4x^3
Divide both sides of the equation by 4 to isolate x^3:
4x^3 = 1.8 x 10^-5 / 4
Simplifying the right side of the equation:
x^3 = 4.5 x 10^-6
To solve for x, take the cube root of both sides:
x = (4.5 x 10^-6)^(1/3)
Calculating this value will give you the molar solubility of Ag2SO4 in water.
...................Ag2SO4 ==> 2Ag^+ + SO4^2-
I...................solid...............0...................0
C..................solid...............2x.................x
E..................solid................2x.................x
Ksp = (Ag^+)^2(SO4^2-)
Substitute the E line into the Ksp expression and solve for x. x is the solubility of Ag2SO4. Post your work if you get stuck.