Posted by **Marcus** on Sunday, March 30, 2008 at 10:59pm.

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.

- chemistry -
**DrBob222**, Sunday, March 30, 2008 at 11:34pm
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.

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