Question

Question #92b85

Answer 1

No, a precipitate will not form.

Explanation:

Start by writing the balanced chemical equation for this double replacement reaction

`"Cr"("NO"_3)_text(3(aq]) + color(red)(3)"NaF"_text((aq]) -> "CrF"_text(3(s]) darr + 3"NaNO"_text(3(aq])`

So, chromium(III) nitrate will react with sodium fluoride to produce chromium(III) fluoride and sodium nitrate if and only if the concentrations of the reactants are in the right proportion.

To determine what that proportion is, use the net ionic equation for this reaction

`"Cr"_text((aq])^(3+) + color(red)(3)"F"_text((aq])^(-) rightleftharpoons "CrF"_text(3(s]) darr`

By definition, the solubility product constant for this reaction will be

`["Cr"^(3+)] * ["F"^(-)]^color(red)(3) = K_(sp)`

In order for a precipitate to form, you need to have

`overbrace(["Cr"^(3+)] * ["F"^(-)]^color(red)(3))^(color(blue)("ion product")) > K_(sp)`

That inequality means that in order for a precipitate to form, the concentrations of the ions must be greater than the equilibrium concentrations.

When the ion product is greater than `K_(sp)`, the solution is supersaturated and so a precipitate will form. If the ion product is smaller than `K_(sp)`, the solution is unsaturated and no precipitate will form.

When the ion product is equal to `K_(sp)`, the solution is saturated. This is the equilibrium point at which no more solid precipitates or dissolves.

Use the molarities and volumes of the two solutions to find how many moles of each you're adding

`color(blue)(c = n/V implies n = c * V)`

For chromium(III) nitrate, you will have

`n = "0.300 M" * 200.0 * 10^(-3)"L" = "0.0600 moles Cr"("NO"_3)_3`

For sodium fluoride, you will have

`n = 4.0 * 10^(-4)"M" * 100.0 * 10^(-3)"L" = 4.00 * 10^(-5)"moles NaF"`

The total volume of the resulting solution will be

`V_"total" = 200.0 + 100.0 = "300.0 mL"`

This means that the concentrations of the two ions in the final solution will be

`["Cr"^(3+)] = "0.0600 moles"/(300.0 * 10^(-3)"L") = "0.200 M"`

`["F"^(-)] = (4.0 * 10^(-5)"moles")/(300.0 * 10^(-3)"L") = 1.33 * 10^(-4)"M"`

So, plug in these values and see if they satisfy the aforementioned inequality

`0.200 * (1.33 * 10^(-4))^color(red)(3) > K_(sp)`

`4.7 * 10^(-13) color(red)(cancel(color(black)(>))) 6.6 * 10^(-11)`

This means that the solution is unsaturated, and a precipitate will not form.