Question

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

Answer 1

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 K_sp 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.