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

Question

Olivia Anton · Accepted Answer

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. Since sodium hydroxide and calcium nitrate are both soluble salts, they will fully separate in an 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 implies that the hydroxide anions and calcium cations will have concentrations that are #["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 For this double replacement reaction, the overall balanced equation is #"Ca"("NO"_3)_text(2(aq]) + color(red)(2)"NaOH"_text((aq]) -> "Ca"("OH")_text(2(s]) darr + 2"NaNO"_text(3(aq])# This is how the entire ionic equation will appear. #"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])^(-)# After eliminating spectator ions, the net ionic equation 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)# bilbo.chm.uri.edu/CHM112/tables/KspTable.htm The solubility product constant is defined as follows by definition: #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 precisely, you must possess #color(blue)(Q_(sp) > K_(sp)) -># a precipitate is formed With the crucial exception of not requiring equilibrium concentrations, the ion product has the same form as the solubility ion product. #Q_(sp) = ["Ca"^(2+)] * ["OH"^(-)]^color(red)(2)# Enter your values to get; for simplicity's sake, I'll omit the units. #Q_(sp) = 0.0175 * (0.0175)^color(red)(2)# #Q_(sp) = 5.36 * 10^(-6)# Given this disparity #Q_(sp) color(red)(cancel(color(black)(>))) K_(sp)# is not true; mixing those two solutions will not result in a precipitate. 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.

Caleb Adams · Answer

undefined Yes, a precipitate will form.