Question

In pure water at 25 °C, the concentration of a saturated solution of CuF2 is 7.4 × 10−3 M. If measured at the same temperature, what is the concentration of a saturated solution of CuF2 in aqueous 0.20 M NaF

Answer 1

The concentration of a saturated solution of CuF₂ in aqueous 0.20 M NaF is 4.0×10⁻⁵ M.

Step-by-step explanation:

Consider the ICE take for the solubility of the solid, CuF₂ as:

CuF₂ ⇄ Cu²⁺ + 2F⁻

At t=0 x - -

At t =equilibrium (x-s) s 2s

The expression for Solubility product for CuF₂ is:

`K_(sp)=\left [ Cu^(2+) \right ]\left [ F^- \right ]^2`

`K_(sp)=s* {2s}^2`

`K_(sp)=4s^3`

Given s = 7.4×10⁻³ M

So, Ksp is:

`K_(sp)=4* (7.4* 10^(-3))^3`

`K_(sp)=4* (7.4* 10^(-3))^3`

Ksp = 1.6209×10⁻⁶

Now, we have to calculate the solubility of CuF₂ in NaF.

Thus, NaF already contain 0.20 M F⁻ ions

Consider the ICE take for the solubility of the solid, CuF₂ in NaFas:

CuF₂ ⇄ Cu²⁺ + 2F⁻

At t=0 x - 0.20

At t =equilibrium (x-s') s' 0.20+2s'

The expression for Solubility product for CuF₂ is:

`K_(sp)=\left [ Cu^(2+) \right ]\left [ F^- \right ]^2`

`1.6209* 10^(-6)={s}'* ({0.20+2{s}'})^2`

Solving for s', we get

s' = 4.0×10⁻⁵ M

The concentration of a saturated solution of CuF₂ in aqueous 0.20 M NaF is 4.0×10⁻⁵ M.