How many moles of baf2 are soluble in 250ml of sodium fluoride (naf)? concentration of naf = 0.12 mol/
pksp of baf2 = 5.8

0.0000278 moles of barium fluoride are soluble in 250ml of sodium fluoride.

Explanation:

Concentration of sodium fluoride (NaF) = 0.12 mol/L

1 mole of NaF gives 1 mole of fluoride ions.

Concentration of fluoride ions from NaF=

Concentration of barium fluoride:

of barium fluoride = 5.8

Solubility product of barium fluoride :

               S      (2S+ 0.12)

Total Concentration of barium ions in a final solution = S

Total Concentration of fluoride ions in a final solution= (2S+ 0.12)

An expression of is given as:

Since, barium fluoride is sparingly soluble salt. Value of S will be small so we can neglect values

Calculating for S;

S = 0.00011 mol/L

According to reaction , 1 mol of barium fluoride gives 1 mole of barium ion.Then 0.00011 barium ions will obtained from 0.00011 mol/L of barium fluoride solution.

So the concentration of barium fluoride is 0.00011 mol/L.

Volume of the solution = 250 ml = 0.250 L

n = 0.0000278 mol

0.0000278 moles of barium fluoride are soluble in 250ml of sodium fluoride.

answer: mole of the gas

answer: mole of the gasexplanation: