calculate the solubility of barium sulfate (Ksp=1.1x10^-10) in
a) water
b) a 0.10 M barium chloride solution
help is GREATLY appreciated
In water: let x = solubility
.......BaSO4 => Ba^2+ + SO4^2-
.........x.......x.......x
Ksp = (Ba^2+)(SO4^2-)
Substitute from the chart above and solve for x =solubility.
For 0.1M BaCl2.
For (Ba^2+) = x from BaSO4 and 0.1 from BaCl2 for a total of 0.1+x
x = SO4^2-
Substitute into Ksp and solve for x = solubility.
The latter is an example of the common ion effect.
To calculate the solubility of barium sulfate (BaSO4), we need to consider the equilibrium expression for the dissolution of BaSO4, also known as the solubility product constant (Ksp). The balanced chemical equation for the dissolution of BaSO4 is:
BaSO4(s) ↔ Ba2+(aq) + SO42-(aq)
Let's calculate the solubility of BaSO4 in both scenarios:
a) Solubility in water:
In pure water, we assume that the initial concentration of both Ba2+ and SO42- ions is zero. Let's assume x moles per liter (M) of BaSO4 dissolves.
The equilibrium expression for the dissolution of BaSO4 is:
Ksp = [Ba2+][SO42-]
Ksp is given as 1.1x10^-10, so we can write:
1.1x10^-10 = x * x
1.1x10^-10 = x^2
Taking the square root of both sides, we find:
x ≈ 1.05x10^-5 M
Therefore, the solubility of BaSO4 in water is approximately 1.05x10^-5 M.
b) Solubility in a 0.10 M barium chloride (BaCl2) solution:
In this case, the presence of the common ion Ba2+ from BaCl2 will affect the solubility of BaSO4.
Since 0.10 M barium chloride is a strong electrolyte, it completely dissociates into Ba2+ and Cl- ions. So, the initial concentration of Ba2+ is 0.10 M.
Now, let's proceed with the calculation using the same equilibrium expression as before:
Ksp = [Ba2+][SO42-]
1.1x10^-10 = (0.10 + x) * x
Since the concentration of Ba2+ is 0.10 M initially, we add it to x to account for the additional amount of Ba2+ coming from the BaCl2 solution.
Simplifying the equation:
1.1x10^-10 = 0.10x + x^2
Rearranging and solving this quadratic equation may seem complicated, but we can make an approximation in this case since the Ksp value is very small compared to the 0.10 M concentration of Ba2+.
By neglecting x in comparison to 0.10 M, we can simplify the equation to:
1.1x10^-10 ≈ 0.10x
Therefore,
x ≈ (1.1x10^-10 / 0.10)
x ≈ 1.1x10^-9 M
So, the solubility of BaSO4 in a 0.10 M barium chloride solution is approximately 1.1x10^-9 M.
Remember that these calculations are approximations, and to get accurate results, more advanced techniques such as the quadratic formula or numerical methods may be required.