# Question 5be33

Apr 6, 2015

!! LONG ANSER !!

SIDE NOTE This in not the correct way to solve this problem, because it assumes that all the volume of the salt will add to that of the calcium chloride solution; this is not correct, like Alfredo showed in his answer and in the later comments.

However, I will leave the answer as it is because I want students to be able to see the mistake I made in solving this problem.

So, here's how you'd go about solving this problem.

You know that you are dealing with a 0.250-M solution of calcium chloride in an initial volume of ${\text{50.0 cm}}^{3}$. Because calcium chloride is soluble in aqueous solution, you know how many moles of calcium cations, $C {a}^{2 +}$, and chloride anions, $C {l}^{-}$, are present in your initial solution.

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

Notice that you have a $1 : \textcolor{red}{2}$ mole ratio between calcium chloride and the chloride ions, which means that your solution will contain twice as many moles of $C {l}^{-}$ as it does moles of $C a C {l}_{2}$.

At the same time, the $1 : 1$ mole ratio that exists between calcium chloride and the calcium cations implies that the number of moles of $C {a}^{2 +}$ is equal to that of $C a C {l}_{2}$.

Use the solution's molarity to determine how many moles of each you get

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

${n}_{C a C {l}_{2}} = {\text{0.350 M" * 50.0 * 10^(-3)"L" = "0.0175 moles Cl}}^{-}$

This means that you get

0.0175cancel("moles CaCl"_2) * "1 mole Ca"^(2+)/(1cancel("mole CaCl"_2)) = "0.0175 moles Ca"^(2+)

and

0.0175cancel("moles CaCl"_2) * "2 moles Cl"^(-)/(1cancel("mole CaCl"_2)) = "0.035 moles Cl"^(-)

Now you add the 2.25 g of sodium chloride. Because the volumes are additive, you must determine what volume that much $N a C l$ will occupy; this is done by using its density, given as $2.1625 {\text{g/cm}}^{3}$

$\rho = \frac{m}{V} \implies V = \frac{m}{\rho}$

V_(NaCl) = (2.25cancel("g"))/(2.1625cancel("g")/"cm"^3) = "1.04 cm"^3

Once you add it to the solution, the number of moles of sodium cations and of chloride ions will be

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

2.25cancel("g") * "1 mole NaCl"/(58.44cancel("g")) = "0.0385 moles NaCl"

0.0385cancel("moles NaCl") * "1 mole Na"^(+)/(1cancel("moles NaCl")) = "0.0385 moles Na"^(+)

and

0.0385cancel("moles NaCl") * "1 mole Cl"^(-)/(1cancel("moles NaCl")) = "0.0385 moles Cl"^(-)

The total volume of the solution will be

V_("total") = V_("NaCl") = V_("initial") = 1.04 + 50.0 = "51.04 cm"^3

It is safe to assume, given the lack of information provided, that the total volume of the solution will not be changed; it is beyond the scope of the given data to show that it does actually change, but by less than 1.04 mL.

The total number of chloride ions will be

${n}_{\text{chloride") = n_("initial") + n_("added}}$

n_("chloride") = 0.035 + 0.0385 = "0.0735 moles Cl"^(-)

Now just solve for each ion's concentration by using the total volume

C_(Na^(+)) = "0.0385 moles"/(51.04 * 10^(-3)"L") = color(green)("0.754 M")

C_(Ca^(2+)) = "0.0175 moles"/(51.04 * 10^(-3)"L") = color(green)("0.343 M")

C_(Cl^(-)) = "0.0735 moles"/(51.04 * 10^(-3)"L") = color(green)("1.44 M")#

Apr 6, 2015

This kind of problem is solved by
1. determining the quantities of each kind of solute and ion, then
2. summing the quantities of ions of the same species, and finally
3. dividing each ion amount by the total volume, to determine the corresponding concentration.

This is best accomplished by converting firstly the quantities of substance in moles.

The problem is badly formulated, because it refers to an additive mixing of volumes of liquid solutions, whereas the mass of 2.25 g of NaCl is supposedly added as a solid to the 50 cm³ of calcium chloride solution.

Anyway I will try to solve the problem by interpreting the "additivity" criterion to mean that the volume of calcium chloride solution will remain approximately 50 cm³ after the addition of 2.25 grams of NaCl.

1. Calculate the amount of each substance in moles
Molar mass of NaCl = $\left(22.99 + 35.45\right) \frac{g}{m o l} = 58.44 \frac{g}{m o l}$;
moles of 2.25 g $N a C l = \frac{2 , 25 \text{ g}}{58.44 \frac{g}{m o l}} = 0.0385 m o l$.

Moles of $C a C {l}_{2} = 0.050 \text{ L" · "0.350 " (mol)/L = 0.0175" mol}$.

2. Calculate the amount of each ion in moles (the two salts are strong electrolytes i.e. completely dissociated in ions)

$w a t e r + 1 N a C l \left(s\right) \to 1 N {a}^{+} \left(a q\right) + 1 C {l}^{-} \left(a q\right)$
$w a t e r + 1 C a C {l}_{2} \left(s\right) \to 1 C {a}^{2 +} \left(a q\right) + 2 C {l}^{-} \left(a q\right)$

Therefore,

${n}_{N {a}^{+}} = {n}_{N a C l} = 0.0385 \text{ moles}$
${n}_{C {l}^{-}} = {n}_{N a C l} + 2 \cdot {n}_{C a C {l}_{2}} = 0.0385 + 2 \cdot 0.0175$ =
$\text{ "= 0.0385+0.035 = 0.0735 " moles}$
${n}_{C {a}^{2 +}} = {n}_{C a C {l}_{2}} = \text{0.0175 } m o l$

Calculate the (molar) concentration of each ion species
${C}_{N {a}^{+}} = \frac{0.0385}{0.050} = 0.77 M$

${C}_{C {l}^{-}} = \frac{0.0735}{0.050} = 1.47 M$

${C}_{C {a}^{2 +}} = \frac{0.0175}{0.050} = 0.35 M$