Question

Given that the k sp for caf2 at 25°c is 3.9 × 10–11, will a precipitate form when 0.10 l of 2.3 × 10–4 m cacl2 is added to 0.50 l of 5.0 × 10–3 m naf? assume volumes are additive.

Answer 1

Given:

Ksp (CaF2) at 25 degrees Celsius = 3.9 x10^-11
volume of CaCl2 = 100 ml
concentration of CaCl2 = 2.3 x 10^-4 m
Volume of NaF = 500 ml
concentration of NaF = 5.0x10^-3 m

Total volume = 100 + 500 = 600 ml

balanced chemical equation:

2NaF + CaCl2 ===> CaF2 + 2NaCl

calculate the number of moles NaF and CaCl2. Then determine the resulting moles of CaF2 (based on the limiting reactant) and find the concentration of CaF2. If it is more than the Ksp, meaning there will be a precipitate.

Answer 2

Answer:- Precipitate will form.

Solution:- A precipitate is formed if the ionic product is greater than solubility product, (
`Q_s_p>K_s_p` )

0.10 L of calcium chloride and 0.50 L of sodium fluoride are added. So, total volume of the solution = 0.10 L + 0.50 L = 0.60 L

We get the calcium ions from calcium chloride as:

`CaCl_2(aq)\rightarrow Ca^2^+(aq)+2Cl^-(aq)`

From this equation, there is 1:1 mol ratio between calcium chloride and calcium ion. So, the initial concentration of calcium ion will be same as of calcium chloride. As the volume is additive, the final concentration of calcium ion is calculated using the dilution equation:

`C_1V_1=C_2V_2`

It could also be written as,
`C_2=(C_1V_1)/(V_2)`

Let's plug in the values in it:

`C_2=(2.3*10^-^3M*0.10L)/(0.60L)`

`C_2=3.8*10^-^4M`

Similarly there is 1:1 mol ratio between sodium fluoride and fluoride ions. we can calculate the final concentration of Fluoride ion as:

`C_2=(5.0*10^-^3M*0.50L)/(0.60L)`

`C_2=4.2*10^-^3M`

Now, we will calculate the ionic product as:

`Q_s_p=[Ca^2^+][F^-]^2`

`Q_s_p=[3.8*10^-^4][4.2*10^-^3]^2`

`Q_s_p=1.6*10^-^9`

Given solubility product is
`3.9*10^-^1^1` which is less than the calculated value of ionic product. (
`Q_s_p>K_s_p` ) .

Hence. the precipitate will form.