Will a precipitate form when we micx Ca(NO3)2(aq) with NaOH(aq) if the concentrations after mixing are both 0.0175M?
1 Answer
Possibly no.
Explanation:
The idea here is that calcium nitrate,
Both calcium nitrate and sodium hydroxide are soluble salts, so they will dissociate completely in aqueous solution to form
#"Ca"("NO"_3)_text(2(aq]) -> "Ca"_text((aq])^(2+) + 2"NO"_text(3(aq])^(-)#
#"NaOH"_text((aq]) -> "Na"_text((aq])^(+) + "OH"_text((aq])^(-)#
This means that the concentrations of the calcium cations and hydroxide anions will be
#["Ca"^(2+)] = 1 xx ["Ca"("NO"_3)_2] -># one mole of calcium nitrate produces one mole of calcium cations
#["OH"^(-)] = 1 xx ["NaOH"] -># one mole of sodium hydroxide produces one mole of hydroxide anions
The overall balanced equation for this double replacement reaction is
#"Ca"("NO"_3)_text(2(aq]) + color(red)(2)"NaOH"_text((aq]) -> "Ca"("OH")_text(2(s]) darr + 2"NaNO"_text(3(aq])#
The complete ionic equation will look like this
#"Ca"_text((aq])^(2+) + 2"NO"_text(3(aq])^(-) + color(red)(2)"Na"_text((aq])^(+) + color(red)(2)"OH"_text((aq])^(-) -> "Ca"("OH")_text(2(s]) darr + 2"Na"_text((aq])^(+) + 2"NO"_text(3(aq])^(-)#
The net ionic equation, for which spectator ions are eliminated, will be
#"Ca"_text((aq])^(2+) + color(red)(2)"OH"_text((aq])^(-) -> "Ca"("OH")_text(2(s]) darr#
Now, the solubility product constant,
#K_(sp) = 5.5 * 10^(-6)#
http://bilbo.chm.uri.edu/CHM112/tables/KspTable.htm
By definition, the solubility product constant, is defined as
#K_(sp) = ["Ca"^(2+)] * ["OH"^(-)]^color(red)(2)#
In order to determine whether or not a precipitate is formed, you need to calculate the ion product,
More specifically, you need to have
#color(blue)(Q_(sp) > K_(sp)) -># a precipitate is formed
The ion product takes the exact same form as the solubility ion product, with the important difference that it does not use equilibrium concentrations.
#Q_(sp) = ["Ca"^(2+)] * ["OH"^(-)]^color(red)(2)#
Plug in your values to get - I'll skip the units for the sake of simplicity
#Q_(sp) = 0.0175 * (0.0175)^color(red)(2)#
#Q_(sp) = 5.36 * 10^(-6)#
Since this inequality
#Q_(sp) color(red)(cancel(color(black)(>))) K_(sp)#
is not valid, a precipitate will not form when you mix those two solutions.
Mind you, the answer depends on the value for