Answer:
BaF2 = Ba2+ + 2F-
Conc s s 2s
Ksp=[Ba2+][F-]2
=[s][2s]2
=4s3
Given, Ksp= 1.84x10-7
So, 4s3 = 1.84x10-7
s3 = 0.46x10-7
s = 3.58x10-3
So, if concentration of Ba2+ ion increases beyond 3.58x10-3 M then it will start precipitating.
Molecular weight of Ba(NO3)2 =261.3 gm/mol
Reaction will progress in solution is BaNO3 + NaF = BaF + NaNO3.
Concentration of F- = 0.0053 mol/L = 5.3X10-3
So, in new solution if Ksp increases above the given value, precipitation start.
1.84x10-7=[s][5.3X10-3]2
s= 0.0065 M
Highest concentration of Ba2+ will be 0.0065 M.
So weight required to make 750 ml 0.065 M Ba(NO3)2 solution is = 0.75 X 261.3 X 0.0065 gm =1.273 gm.
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