# chemistry

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Calculate the xolubility of calcium sulfate in 0.010 mol/L calcium nitrate at SATP

CaSO4 ==> Ca^+2 + SO4^=
Ca(NO2)2 ==> Ca^+2 + 2NO3^- = 0.01 M
Ksp CaSO4 = (Ca^+2)(SO4^=) = ??. Look up Ksp for CaSO4.

Subsitute 0.01 for Ca^+2 and x for SO4^= and solve for solubility CaSO4. Post your work if you get stuck.

Ksp is 7.1 x 10^-5 for CaSO4

7.1x 10^-5 = (0.01)(x)

x = 0.000071/0.01

=0.0071

Why am I wrong?

• chemistry -

(Ca^+2) = x from CaSO4 and 0.01 from Ca(NO3)2.
(SO4^=) = x
(0.01+x)(x)=7.1 x 10^-5
When you solved for x, the value of 0.0071 you have is correct; however, in relation to 0.01, it is too large. In other words, the total Ca^+2 is 0.01 + solubility of CaSO4. If the solubility (in this case x) is larger than about 5% of the common ion (0.01 in this case), you can't make the simplifying assumption that 0.01+x = 0.01. For example, you found 0.0071 for the solubility and that + 0.01 = 0.0171 which is MUCH MUCH greater than 5%. Therefore, a large error is made by making the simplifying assumption. What does that mean for you (and the problem)? It must be solved by the quadratic formula.
(x+0.01)(x) = 7.1 x 10^-5
x^2 + 0.01x -7.1 x 10^-5 = 0
Make sure that the value of Ksp you are using is from the same source as your problem for if you are using a different Ksp than the person who put all those numbers in the data base you are checking then you may NEVER get the right answer. To let you know if your solution is correct or not, assuming Ksp really is 7.1 x 10^-5 (its about 5 something x 10^-5 in my references), I find solubility = 0.0048 M. Let me know if you have trouble.

• chemistry -

thank you

• chemistry -

i beat lindan in previous china open superseries hurray

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