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QUESTION

# Will a precipitate form when we micx Ca(NO3)2(aq) with NaOH(aq) if the concentrations after mixing are both 0.0175M?

Possibly no.

The idea here is that calcium nitrate, "Ca"("NO"_3)_2, and sodium hydroxide, "NaOH", will react to form calcium hydroxide, an insoluble solid, if and only if they are mixed in the appropriate concentrations.

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 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), for calcium hydroxide is listed as being equal to

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, Q_(sp), for this reaction.

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 .

Mind you, the answer depends on the value for K_(sp) given to you by the problem. Since you didn't provide one here, you will have to compare the value of Q_(sp) with the value of K_(sp) given to you, and say if a precipitate will form or not.