Calculate the solubility of MgSO4(Ksp=1.0*10^-10 at 298 K) in: (a) pure water (b)1.0M Na2SO4. (Express solubilities in mol/L).

Asked by verma2 | 23rd Jan, 2012, 11:33: PM

Expert Answer:

MgSO4 (s) <---> Mg2+ (aq) + SO42- (aq)
 
Ksp = [Mg2+][SO42-] = 1.0 x 10-10 at 298 K
 
We are not given any initial concentration of any of these ions so we say that Mg2+ have 0 M (moles/liter) and SO42- have 0 M. However, since the solid magnisium sulfate dissolves to some extent and we do not know the number (which is what we are trying to figure out), we say that X amount has been added to both ions. So the resulting concentration in equilibrium is [Mg2+] = 0 + X = X and [SO42- ] = 0 + X = X.
 
Since, [Mg2+] = [SO42-] = x
 
we can write [Mg2+]2 = x2  = 1.0 x 10-10 
 
or  [Mg2+] = (1.0 x 10-10 )1/2 = 1.0 x 10-5
 
 
 
 

Answered by  | 24th Jan, 2012, 10:44: AM

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