Molar sol. from Ksp

[tutankh]

This is problem #6, Passgae 2-Gen Chem, MCAT official guide-page 97.

I calculated the molar solubility as follows:

x^2 = Ksp(CaCo3) = 3.3 x 10^-9

x = √3.3 x 10^-9

According to the answer key, the correct answer is (A), which is 5.5 x 10^-5 M

I got: 1.8-someting x 10^-5

Am I missing anything here?

---

Members don't see this ad.

 

[inaccensa]

This is problem #6, Passgae 2-Gen Chem, MCAT official guide-page 97.

I calculated the molar solubility as follows:

x^2 = Ksp(CaCo3) = 3.3 x 10^-9

x = √3.3 x 10^-9

According to the answer key, the correct answer is (A), which is 5.5 x 10^-5 M

I got: 1.8-someting x 10^-5

Am I missing anything here?

We know that 3.3 is not a good no like lets say 2.5 or 1.6. What you can do here is 3.3*10^-9 =33*10^-10. Now we know that the sq rt of 10^-10 will be 10^-5. What abt 33? Think of a no whose square is close to 33. sq of 6 =36 and 5 =25. Since the no is 33, it will be between 5 and 6, but very close to 6