Question

3) a) Sodium fluoride has a K_sp of 0.934. If 15.00 g of sodium fluoride is placed in a beaker containing 250. mL of water, what are the concentrations of the ions in solution after the solution is at equilibrium?

b) If solid in (a) was placed in 250. mL of a 0.200 M NaOH solution instead of pure water, what mass of sodium fluoride would dissolve

Answer 1

3.

(a)

                                                              NaF         Na⁺             +    F^-

Initial concentration                                                             0                                    0

Change in concentration                       -Y                          Y                                    Y

Equilibrium concentration                      -Y                          Y                                    Y

K_sp = [Na+][F-]

K_sp = (Y)(Y)

or Y² = K_sp = 0.934

or, Y = 0.966

Therefor the molar solubility of NaF = 0.966 M

Molar mass of NaF = 42 g/mol

Therefore [NaF] = 42 x 0.966 g/L = 40.572 g/L

This means 1 L water dissolves 40.572 g NaF

Therfore 250 mL or 0.25 L water dissolves 40.572 x 0.25 = 10.143 g

[Na⁺] = 23 x 0.966 g/L = 22.218 g/L             ( Molar mass of Na⁺ = 23 g/mol)

Therefore the amount of Na⁺ in 0.25 L = 22.218 x 0.25 = 5.5545 g

[F^-] = 19 x 0.966 g/L = 18.354 g/L                 ( Molar mass of F^- = 19 g/mol)

Therefore the amount of F^- in 0.25 L = 18.354 x 0.25 = 4.5885 g

Therefore the concentration of Na⁺ = 5.5545 g/L

And the concentration of F^- = 4.5885 g/L

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(b)

When the solid was placed in 250 mL of 0.200 M NaOH, the initial [Na+] = 0.200 M

                                                               NaF         Na⁺             +    F^-

Initial concentration                                                             0.200                             0

Change in concentration                       -Y                     (0.200+Y)                           Y

Equilibrium concentration                      -Y                     (0.200+Y)                            Y

K_sp = [Na+][F-]

K_sp = (0.200+Y)(Y)

or Y² + 0.200Y = K_sp = 0.934

or, Y² + 0.200Y - 0.934 = 0

or,

or,

or,

or,

Y = 0.872      (Negative value is discarded)

Therefor the molar solubility of NaF = 0.872 M

Molar mass of NaF = 42 g/mol

Therefore [NaF] = 42 x 0.872 g/L = 36.624 g/L

This means 1 L water dissolves 36.624 g NaF

Therfore 250 mL or 0.25 L water dissolves 36.624 x 0.25 = 9.156 g

[Na⁺] = 23 x 0.872 g/L = 20.056 g/L             ( Molar mass of Na⁺ = 23 g/mol)

Therefore the amount of Na⁺ in 0.25 L = 20.056 x 0.25 = 5.014 g

[F^-] = 19 x 0.872 g/L = 16.568 g/L                ( Molar mass of F^- = 19 g/mol)

Therefore the amount of F^- in 0.25 L = 16.568 x 0.25 = 4.142 g

Therefore the concentration of Na⁺ = 5.014 g/L

And the concentration of F^- = 4.142 g/L