Answer to Question #101639 in General Chemistry for Ashna

Question #101639
calculate the maximum possible mass of barium nitrate that can be added to a 750ml of 0.0053mol/L sodium fluoride before a precipitate will form. The ksp of barium fluoride is 1.84x10^-7
1
Expert's answer
2020-01-24T05:00:35-0500

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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