Question

An aqueous solution of sodium fluoride is slowly added to a water sample that contains barium ion (2.75×10^-2M ) and calcium ion (7.30×10^-2M ). What is the remaining concentration of the first ion to precipitate when the second ion begins to precipitate?

Answer 1

We must know which ion will precipitate first. To find out this, we need to calculate minimum concentration of Fluoride ions required to start precipitation of BaF ₂ and CaF₂ .

Consider dissociation of BaF₂ in water , BaF ₂ (s) Ba ²⁺ (aq) + 2 F ^- (aq)

For above reaction, Ksp = [ Ba ²⁺ ][F ^- ] ² = 1.0 $\times$ 10 ^-06

$\therefore$ [F ^- ] minimum = ( 1.0 $\times$ 10 ^-06 / [ Ba ²⁺ ] ) ^1/2

[F ^- ] minimum = ( 1.0 $\times$ 10 ^-06 / 2.75 $\times$ 10 ^-02 ) ^1/2

[F ^- ] minimum = 6.030 $\times$ 10 ^-03 M

Consider dissociation of CaF₂ in water , CaF ₂ (s) Ca ²⁺ (aq) + 2 F ^- (aq)

For above reaction, Ksp = [ Ca ²⁺ ][F ^- ] ² = 5.3 $\times$ 10 ^-09

$\therefore$ [F ^- ] minimum = ( 5.3 $\times$ 10 ^-09 / [ Ca ²⁺ ] ) ^1/2

[F ^- ] minimum = ( 5.3 $\times$ 10 ^-09 / 7.30 $\times$ 10 ^-02 ) ^1/2

[F ^- ] minimum =2.694 $\times$ 10 ^-04 M

From above values , it is clear that Calcium fluoride will precipitate out first.

We have to find out [ Ca ²⁺ ] when BaF₂ starts precipitates out.

When BaF₂ starts precipitates out , [F ^- ] is 6.030 $\times$ 10 ^-03 M ( $\because$ [F ^- ] minimum = 6.030 $\times$ 10 ^-03 M)

$\therefore$ [ Ca ²⁺ ] = K sp / [F ^- ] = 5.3 $\times$ 10 ^-09 /  6.030 $\times$ 10 ^-03 = 8.79 $\times$ 10 ^-07 M .

ANSWER : [ Ca ²⁺ ] when BaF₂ starts precipitates out is 8.79 $\times$ 10 ^-07 M .