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

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

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Answer 1 · 2016-02-01T16:20:43

Possibly no.

Explanation:

The idea here is that calcium nitrate, $Ca {({NO}_{3})}_{2}$ $"Ca"("NO"_3)_2$ , and sodium hydroxide, $NaOH$ $"NaOH"$ , 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

$Ca {({NO}_{3})}_{2(aq]} \to {Ca}_{(aq]}^{2 +} + 2 {NO}_{3(aq]}^{-}$ $"Ca"("NO"_3)_text(2(aq]) -> "Ca"_text((aq])^(2+) + 2"NO"_text(3(aq])^(-)$

${NaOH}_{(aq]} \to {Na}_{(aq]}^{+} + {OH}_{(aq]}^{-}$ $"NaOH"_text((aq]) -> "Na"_text((aq])^(+) + "OH"_text((aq])^(-)$

This means that the concentrations of the calcium cations and hydroxide anions will be

$[{Ca}^{2 +}] = 1 \times [Ca {({NO}_{3})}_{2}] \to$ $["Ca"^(2+)] = 1 xx ["Ca"("NO"_3)_2] ->$ one mole of calcium nitrate produces one mole of calcium cations

$[{OH}^{-}] = 1 \times [NaOH] \to$ $["OH"^(-)] = 1 xx ["NaOH"] ->$ one mole of sodium hydroxide produces one mole of hydroxide anions

The overall balanced equation for this double replacement reaction is

$Ca {({NO}_{3})}_{2(aq]} + 2 {NaOH}_{(aq]} \to Ca {(OH)}_{2(s]} ⏐ ⏐ ↓ + 2 {NaNO}_{3(aq]}$ $"Ca"("NO"_3)_text(2(aq]) + color(red)(2)"NaOH"_text((aq]) -> "Ca"("OH")_text(2(s]) darr + 2"NaNO"_text(3(aq])$

The complete ionic equation will look like this

${Ca}_{(aq]}^{2 +} + 2 {NO}_{3(aq]}^{-} + 2 {Na}_{(aq]}^{+} + 2 {OH}_{(aq]}^{-} \to Ca {(OH)}_{2(s]} ⏐ ⏐ ↓ + 2 {Na}_{(aq]}^{+} + 2 {NO}_{3(aq]}^{-}$ $"Ca"_text((aq])^(2+) + 2"NO"_text(3(aq])^(-) + color(red)(2)"Na"_text((aq])^(+) + color(red)(2)"OH"_text((aq])^(-) -> "Ca"("OH")_text(2(s]) darr + 2"Na"_text((aq])^(+) + 2"NO"_text(3(aq])^(-)$

The net ionic equation, for which spectator ions are eliminated, will be

${Ca}_{(aq]}^{2 +} + 2 {OH}_{(aq]}^{-} \to Ca {(OH)}_{2(s]} ⏐ ⏐ ↓$ $"Ca"_text((aq])^(2+) + color(red)(2)"OH"_text((aq])^(-) -> "Ca"("OH")_text(2(s]) darr$

Now, the solubility product constant, $K_{s p}$ $K_(sp)$ , for calcium hydroxide is listed as being equal to

$K_{s p} = 5.5 \cdot 10^{- 6}$ $K_(sp) = 5.5 * 10^(-6)$

http://bilbo.chm.uri.edu/CHM112/tables/KspTable.htm

By definition, the solubility product constant, is defined as

$K_{s p} = [{Ca}^{2 +}] \cdot {[{OH}^{-}]}^{2}$ $K_(sp) = ["Ca"^(2+)] * ["OH"^(-)]^color(red)(2)$

In order to determine whether or not a precipitate is formed, you need to calculate the ion product, $Q_{s p}$ $Q_(sp)$ , for this reaction.

More specifically, you need to have

$Q_{s p} > K_{s p} \to$ $color(blue)(Q_(sp) > K_(sp)) ->$ 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.

$Q_{s p} = [{Ca}^{2 +}] \cdot {[{OH}^{-}]}^{2}$ $Q_(sp) = ["Ca"^(2+)] * ["OH"^(-)]^color(red)(2)$

Plug in your values to get - I'll skip the units for the sake of simplicity

$Q_{s p} = 0.0175 \cdot {(0.0175)}^{2}$ $Q_(sp) = 0.0175 * (0.0175)^color(red)(2)$

$Q_{s p} = 5.36 \cdot 10^{- 6}$ $Q_(sp) = 5.36 * 10^(-6)$

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 $K_(sp)$ 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.

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

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