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?
(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.
11 years later and im making the same mistake
thank you
i beat lindan in previous china open superseries hurray
The textbook says it’s 3.6 x 10^-3, any clue as to how they got that answer? I got 4.8 x 10^-3 with the quadratic formula and 7.1 x 10 ^-3 with simplifying too.
Well, it seems like you made a slight calculation mistake there. Let's go through it again:
You correctly started with the equation: Ksp = [Ca^+2][SO4^=] = 7.1 x 10^-5.
You substituted 0.01 M for [Ca^+2] and x for [SO4^=].
So we have: 7.1 x 10^-5 = (0.01)(x).
Now, if we solve for x, we get: x = 7.1 x 10^-5 / 0.01.
By dividing, we find that x = 7.1 x 10^-3.
Therefore, the solubility of calcium sulfate in 0.010 mol/L calcium nitrate at SATP is 7.1 x 10^-3 M, not 0.0071 M as you initially calculated.
Keep in mind that calculations can sometimes be tricky, and mistakes happen to all of us. Don't worry, though – even the best of us get it wrong sometimes!
Your calculation looks correct, but there seems to be a calculation error in the final step.
To calculate the solubility of calcium sulfate (CaSO4) in a 0.010 mol/L calcium nitrate (Ca(NO3)2) solution, you need to use the solubility product constant (Ksp) for calcium sulfate. You correctly looked up the Ksp value for calcium sulfate, which is 7.1 x 10^-5 mol^2/L^2.
The equation for the dissociation of calcium sulfate in water is:
CaSO4 ⇌ Ca^2+ + SO4^2-
Let's call the solubility of calcium sulfate (x), so the concentration of Ca^2+ (from calcium nitrate) is 0.010 mol/L, and the concentration of SO4^2- is also x, as they both come from the dissociation of calcium sulfate.
Now we can set up the Ksp expression for calcium sulfate:
Ksp = [Ca^2+][SO4^2-] = (0.010)(x)
Substituting the known values:
7.1 x 10^-5 = (0.010)(x)
Now, let's solve for x, the solubility of calcium sulfate:
x = (7.1 x 10^-5) / (0.010)
Calculating:
x = 7.1 x 10^-5 / 0.010
x = 7.1 x 10^-3 moles per liter (mol/L)
So, the solubility of calcium sulfate in a 0.010 mol/L calcium nitrate solution at SATP is 7.1 x 10^-3 mol/L.
Please recheck your calculations to identify any errors.