# Question b3e90

Apr 10, 2015

!! LONG ANSWER !!

Here's how you'd go about solving this problem.

Both silver nitrate and calcium chloride are soluble in aqueous solution, but when mixed together will form a precipitate, silver chloride, $A g C l$, and another soluble compound, calcium nitrate, $C a {\left(N {O}_{3}\right)}_{2}$.

$A g N {O}_{3 \left(a q\right)} \to A {g}_{\left(a q\right)}^{+} + N {O}_{3 \left(a q\right)}^{-}$

$C a C {l}_{2 \left(a q\right)} \to C {a}_{\left(a q\right)}^{2 +} + 2 C {l}_{\left(a q\right)}^{-}$

Notice that 1 mole of silver nitrate produces 1 mole of $A {g}^{+}$ and 1 mole of $N {O}_{3}^{-}$ ions, while 1 mole of calcium chloride produces 1 mole of $C {a}^{2 +}$ and 2 moles of $C {l}^{-}$ ions.

Use the molarities of the two solutions to determine how many moles of each ion you get

$C = \frac{n}{V} \implies n = C \cdot V$

${n}_{A g N {O}_{3}} = \text{0.500 M" * 100.0 * 10^(-3)"L" = "0.0500 moles }$ $A g N {O}_{3}$

${n}_{C a C {l}_{2}} = \text{0.500 M" * 100.0 * 10^(-3)"L" = "0.0500 moles }$ $C a C {l}_{2}$

This corresponds to

${n}_{A {g}^{+}} = {n}_{A g N {O}_{3}} = \text{0.0500 moles }$ $A {g}^{+}$
${n}_{N {O}_{3}^{-}} = {n}_{A g N {O}_{3}} = \text{0.0500 moles }$ $N {O}_{3}^{-}$
${n}_{C {a}^{2 +}} = {n}_{C a C {l}_{2}} = \text{0.0500 moles }$ $C {a}^{2 +}$
${n}_{C {l}^{-}} = 2 \cdot {n}_{C a C {l}_{2}} = \text{0.100 moles }$ $C {l}^{-}$

The balanced chemical equation for this double replacement reaction will be

$2 A g N {O}_{3 \left(a q\right)} + C a C {l}_{2 \left(a q\right)} \to 2 A g C {l}_{\left(s\right)} + C a {\left(N {O}_{3}\right)}_{2 \left(a q\right)}$

The complete ionic equation will be

$2 A {g}_{\left(a q\right)}^{+} + 2 N {O}_{3 \left(a q\right)}^{-} + C {a}_{\left(a q\right)}^{2 +} + 2 C {l}_{\left(a q\right)}^{-} \to 2 A g C {l}_{\left(s\right)} + C {a}_{\left(a q\right)}^{2 +} + 2 N {O}_{3 \left(a q\right)}^{-}$

The net ionic equation for this reaction looks like this

$A {g}_{\left(a q\right)}^{+} + C {l}_{\left(a q\right)}^{-} \to A g C {l}_{\left(s\right)}$

The important thing to notice here is that $C {a}^{2 +}$ and $N {O}_{3}^{-}$ are spectator ions, which means that the number of moles of each present in the final solution will be equal to the number of moles of each added when the two solutions are mixed.

Since the $A {g}^{+}$ and $C {l}^{-}$ ions form the precipitate, the number of moles of each present in the final solution will not be equal to the number of moles of each added.

The volume of the final solution will be

V_"total" = V_("AgNO_3) + V_(CaCl_2)
${V}_{\text{total" = 100.0 + 100.0 = "200.0 mL}}$

Now, according to the net ionic equation, the silver cations and the chloride anions react in a $1 : 1$ mole ratio; since you have twice as many moles of $C {l}^{-}$ available to react, $A {g}^{+}$ will act as a limiting reagent.

In other words, all the $A {g}^{+}$ will be consumed, readucing the number of $C {l}^{-}$ moles by half in the process.

n_(Cl^(-)"remaining") = n_(Cl^(-)) - n_(Ag^(+)) = 0.100 - 0.0500 = "0.0500 moles"

${n}_{A {g}^{+} \text{remaining}} = 0$

Therefore, the concentrations of the ions present in solution will be

${C}_{A {g}^{+}} = 0$
C_(Ca^(2+)) = n/V_"total" = "0.0500 moles"/(200 * 10^(-3)"L") = color(green)("0.250 M")#

${C}_{N {O}_{3}^{-}} = {C}_{C {a}^{2 +}} = \textcolor{g r e e n}{\text{0.250 M}}$

${C}_{C {l}^{-}} = {C}_{C {a}^{2 +}} = \textcolor{g r e e n}{\text{0.250 M}}$

Apr 10, 2015

$\left[A {g}_{\left(a q\right)}^{+}\right] = 0$
$\left[N {O}_{3 \left(a q\right)}^{-}\right] = 0.25 \text{mol/l}$
$\left[C {a}_{\left(a q\right)}^{2 +}\right] = 0.25 \text{mol/l}$
$\left[C {l}_{\left(a q\right)}^{-}\right] = 0.25 \text{mol/l}$

The overall equation is:

$2 A g N {O}_{3 \left(a q\right)} + C a C {l}_{2 \left(a q\right)} \rightarrow 2 A g C {l}_{\left(s\right)} + C a {\left(N {O}_{3}\right)}_{2 \left(a q\right)}$

The ionic equation is:

$A {g}_{\left(a q\right)}^{+} + C {l}_{\left(a q\right)}^{-} \rightarrow A g C {l}_{\left(s\right)}$

$c = \frac{n}{v}$

So $n = c \times v$

The number of moles of $A {g}^{+} = 0.5 \times 100 = 50 \text{mmol}$

The number of moles of $C {l}^{-} = 0.5 \times 100 \times 2 = 100 \text{mmol}$

This is because there are 2 moles of $C {l}^{-}$ per mole of $C a C {l}_{2}$

So after the $C {l}^{-}$ ions have reacted with the $A {g}^{+}$ there will be an XS moles of $C {l}^{-}$ equal to $100 - 50 = 50 \text{mmol}$

The $C {a}^{2 +}$ ions are spectators so the number of $C {a}^{2 +}$ ions $= 100 \times 0.5 = 50 \text{mmol}$

After reaction all the $A {g}^{+}$ ions have formed the precipitate so the no. moles $A {g}_{\left(a q\right)}^{+} = 0$

The nitrate(V) ions are spectators so the no. moles $N {O}_{3 \left(a q\right)}^{-}$ remaining $= 100 \times 0.5 = 50 \text{mmol}$

The total volume $= 100 + 100 = 200 \text{ml}$

So to find the final concentration of each ion we divide the number of moles by the total volume, noting that $200 \text{ml"=200xx10^(-3)"L}$

$\left[A {g}_{\left(a q\right)}^{+}\right] = 0$ - they have all formed the precipitate.

$\left[N {O}_{3 \left(a q\right)}^{-}\right] = \frac{50 \times {10}^{- 3}}{200 \times {10}^{- 3}} = 0.25 \text{mol/l}$

$\left[C {a}_{\left(a q\right)}^{2 +}\right] = \frac{50 \times {10}^{- 3}}{200 \times {10}^{- 3}} = 0.25 \text{mol/l}$

$\left[C {l}_{\left(a q\right)}^{-}\right] = \frac{50 \times {10}^{- 3}}{200 \times {10}^{- 3}} = 0.25 \text{mol/l}$