SrCO3 ==> Sr^+2 + CO3^=
write the Ksp expression.
Substitute 4.0 x 10^-5 M. Solve for Ksp.
Ksp = 4.0E-5 = (x)(3x)
is the (3x) squared ??
I think you are having a tough time understanding the concept of solubility product. Perhaps you need to read some more in your text/notes.
SrCO3 ==> Sr^+2 + CO3^=
Ksp = (Sr^+2)(CO3^=) = ??
Solubility from the problem of SrCO3 = 4 x 10^-5 M and since 1 mol SrCO3 gives 1 mol Sr^+2 and 1 mol SO4^=, then those ions = 4 x 10^-5 M also. So
(Sr^+2) = 4 x 10^-5 M
(CO3^=) = 4 x 10^-5 M
Ksp = (4 x 10^-5)(4 x 10^-5) = ??
I have no idea where you obtained x and 3x OR why you asked the question about cubing something. There is nothing to cube BECAUSE there is no coefficient of 3 anywhere in the equation.
where i am pulling this is if you pull up google and type in molar solubility the first website shows you how to do it... that is what i am following and it shows me that in the book too
If you want to type in the web address I will look at it but I know what Ksp is. I'm sure those sites, and your text, are correct. I'm also relatively sure that you are misinterpreting it. Let me cite the solubility product principle to(for) you.
The product of the molar solubility of the ions of a slightly soluble material, each ion raised to a power indicated by its coefficient in the dissociation reaction, is equal to a constant, called solubility product constant, at a constant temperature.
Here is a site on SrCO3^= that I found. (Broken Link Removed)
well i do understand what you did to solve the problem... and i did find the answer... its just that all this stuff im looking at is sqrt everything thats all and i didnt see you do that
When you are looking for the solubility when Ksp is given, square roots and cube roots (sometimes fourth and fifth roots) are the order of the day. But when we are given the solubility, as in SrCO3, then we simply multiply (4x10^-5)(4x10^-5) and its just the multiplication of those two numbers or I could have written it as
(4.0 x 10^-5)^2 = ??
that makes better sense now
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