What is the solubility in grams per liter of SrSO4, in 0.23 M Na2SO4?
Ksp for SrSO4 = 2.5E-7
To determine the solubility of SrSO4 in 0.23 M Na2SO4, we need to use the concept of common ion effect.
The common ion effect states that the solubility of a slightly soluble salt is reduced when a common ion is already present in the solution. In this case, Na2SO4 dissociates into two sodium ions (2Na+) and one sulfate ion (SO4^2-).
Given that the solubility product constant (Ksp) for SrSO4 is 2.5E-7, we can write the equilibrium expression as:
Ksp = [Sr^2+][SO4^2-]
However, in the presence of Na2SO4, there will also be sulfate ions (SO4^2-) from the dissociation of Na2SO4. This means that the concentration of sulfate ions in the solution will be higher than just the concentration from the SrSO4 dissociation.
To account for this, we need to calculate the concentration of sulfate ions in the solution due to Na2SO4 before we can determine the solubility of SrSO4.
The concentration of sulfate ions due to Na2SO4 is given by the molarity of Na2SO4 multiplied by the number of sulfate ions produced per formula unit of Na2SO4. In this case, there is one sulfate ion, so we multiply the molarity of Na2SO4 by 1.
Concentration of SO4^2- due to Na2SO4 = 0.23 M x 1 = 0.23 M
Now, we can write the equilibrium expression for the solubility of SrSO4, taking into account the concentration of sulfate ions from both SrSO4 and Na2SO4:
Ksp = [Sr^2+][SO4^2-] = (x)(0.23 + x)
Since the solubility of SrSO4 is given in grams per liter, we need to convert it to moles per liter in order to substitute it into the equilibrium expression.
Using the molar mass of SrSO4 (183.7 g/mol), we can calculate the number of moles per liter (M):
(x g/L) / (183.7 g/mol) = x mol/L
Now, we can substitute the expression for concentration in moles per liter into the equilibrium expression:
Ksp = (x)(0.23 + x) = 2.5E-7
Solving this quadratic equation will give us the solubility (x) of SrSO4 in moles per liter. We can then convert this value to grams per liter by multiplying by the molar mass of SrSO4.
Once you have solved the equation, you can find the solubility of SrSO4 in grams per liter by multiplying the solubility in moles per liter by the molar mass of SrSO4.
Remember to account for significant figures, units, and any rounding conventions specified in your specific instructions or experiment.
To determine the solubility of SrSO4 in 0.23 M Na2SO4, we need to use the common ion effect. The common ion effect states that adding a common ion to a solution will reduce the solubility of a compound.
In this case, the common ion is sulfate (SO4²¯) from Na2SO4.
Let's assume that x grams of SrSO4 dissolve in 1 liter of solution. So the concentration of Sr²⁺ ions will also be x M.
The concentration of sulfate ions (SO4²¯) in the solution will be the sum of Na2SO4 and SrSO4:
[SO4²¯]total = [SO4²¯]Na2SO4 + [SO4²¯]SrSO4
[SO4²¯]total = 0.23 M + x M
The solubility product expression for SrSO4 can be written as:
Ksp = [Sr²⁺][SO42-]
Substituting the values:
2.5E-7 = x * (0.23 M + x)
To solve this quadratic equation, we can assume that x is much smaller than 0.23 M. This approximation will simplify the calculation.
Therefore, we can ignore x when compared to 0.23 M, resulting in:
2.5E-7 = x * 0.23 M
Simplifying:
x = 2.5E-7 / 0.23
Calculating:
x ≈ 1.09E-6 M
Finally, to convert from molarity to grams per liter, multiply by the molar mass of SrSO4:
Mass = 1.09E-6 M * (g/mol SrSO4) * (L/1 M)
Using the molar mass of SrSO4 (183.68 g/mol):
Mass ≈ 1.09E-6 * 183.68 g/L
Calculating:
Mass ≈ 2.00E-4 g/L
Therefore, the solubility of SrSO4 in 0.23 M Na2SO4 is approximately 2.00E-4 grams per liter.