Please calculate the solubility of calcium sulfate in 0.010 mol/L calcium nitrate at SATP.
This very question has been bugging me all weekend. Can some one please solve this quesiton it would really be appreciated.
CaSO4 ==> Ca^+2 + SO4^=
If ionic strength is to be neglected, and there is no acidity listed (acidity would increase the solubility due to the formation of HSO4^-) then (Ca^+2) = x+0.010, (SO4^=) = x
Ksp = (Ca^+2)(SO4^=) = 2.6 x 10^-5
(x+0.01)(x) = 2.6 x 10^-5
If we call x + 0.01 = x, then
x = 2.6 x 10^-3 = 0.0026 M = (CaSO4).
To see if x + 0.01 = 0.01, we check that 0.01 + 0.0026 = 0.0126 so x really isn't negligible.(or another way is to check the percentage of [0.0026/0.01]*100 = 26% which is too large an error to introduce). You can solve the quadratic but the easier way is to solve by iteration (successive approximations); i.e., use 0.0126 for (Ca^+2) and not 0.01, then
(0.0126)(x) = 2.6 x 10^-5
x = 0.00206. That gives a new (Ca^+) = 0.01 + 0.00206 = 0.01206, then
(0.01206)(x) = 2.6 x 10^-5
x = 0.00216 and 0.01 + 0.00216 = 0.01216 = (Ca^+2).
(0.01216)(x) = 2.6 x 10-5
x = 0.00214 so I would stop there and call the solubility 0.00214 mols/L. You note that 0.00214 is significantly different from the first value of 0.0026 so the 0.00214 M is the better value. If you wish to solve the quadratic you will have
(x+0.01)(x) = 2.6 x 10^-5
X^2 + 0.01x - 2.6 x 10^-5
Solving the quadratic gives 0.00214 for x (and may be easier than successive approximations. USUALLY, approximations is a much easier and faster way of doing it. I hope this helps. It may be more than you ever wanted to know about the problem.
To calculate the solubility of calcium sulfate in 0.010 mol/L calcium nitrate at SATP (standard ambient temperature and pressure), you need to use the solubility product constant (Ksp) and the stoichiometry of the reaction.
The solubility product constant (Ksp) is the equilibrium constant representing the dissolution of a solid compound into its constituent ions in a solution. For calcium sulfate (CaSO4), the dissociation can be represented by the equation:
CaSO4 (s) ⇌ Ca2+ (aq) + SO42- (aq)
The Ksp expression for this dissociation is:
Ksp = [Ca2+] [SO42-]
The given concentration of calcium nitrate (Ca(NO3)2) does not affect the Ksp of calcium sulfate. Therefore, we can consider the concentration of Ca2+ ions supplied by calcium nitrate to be negligible compared to the concentration of Ca2+ ions from calcium sulfate.
Now, to determine the solubility of calcium sulfate in 0.010 mol/L calcium nitrate, we need the Ksp value for calcium sulfate. The Ksp value for calcium sulfate at SATP is approximately 4.93 x 10^(-5) mol^2/L^2.
Using this Ksp value, we can set up the following equation:
4.93 x 10^(-5) = [Ca2+] [SO42-]
Since the calcium sulfate dissociates in a 1:1 ratio, the concentration of Ca2+ is equal to the concentration of SO42-. Let's call the solubility of calcium sulfate x. Therefore:
x = [Ca2+] = [SO42-]
Now, substituting this value back into the equation, we get:
4.93 x 10^(-5) = x^2
Taking the square root of both sides, we find:
x = sqrt(4.93 x 10^(-5))
Calculating this expression, we get:
x ≈ 0.0070 mol/L
So, the solubility of calcium sulfate in 0.010 mol/L calcium nitrate at SATP is approximately 0.0070 mol/L.