Question

Determine the concentrations of BaBr2, Ba2+, and Br− in a solution prepared by dissolving 2.38 × 10−4 g BaBr2 in 2.00 L of water. Express all three concentrations in molarity. Additionally, express the concentrations of the ionic species in parts per million (ppm).

Answer 1

The BaBr₂ will dissolve in water and form its ions according to this equation:

BaBr₂ ----> Ba²⁺ + 2 Br⁻

We have to find the molarity of the solution. The molarity of a solution is calculated using this formula:

Molarity = moles of solute / volume of solution in L

We know the volume of solution (2.00 L) and we know the mass of BaBr₂. To find the molarity of the solution we will have to convert those grams into moles. When we have to convert a mass into moles we use the molar mass. Let's start finding the molar mass of BaBr₂.

atomic mass of Ba: 137.33 amu

atomic mass of Br: 79.90 amu

molar mass of BaBr₂ = 137.33 + 2 * 79.90

molar mass of BaBr₂ = 297.13 g/mol

Now we can find the number of moles of BaBr₂ that we have in 2.38 *10^(-4) g of it.

moles of BaBr₂ = 2.38 *10⁻⁴ g / (297.13 g/mol)

moles of BaBr₂ = 8.01 *10⁻⁷ moles

So we prepared a solution dissolving 8.01 * 10⁻⁷ moles of BaBr₂ in 2.00 L of water. Let's find the molarity:

Molarity of BaBr₂ = moles of BaBr₂ / volume of solution in L

Molarity of BaBr₂ = 8.01 * 10⁻⁷ moles/(2.00 L)

Molarity of BaBr₂ = 4.00 * 10⁻⁷ M

We said that the BaBr₂ will dissolve in water according to this equation:

BaBr₂ ----> Ba²⁺ + 2 Br⁻

If we look at the coefficients we see that 1 mol of BaBr₂ will produce 1 mol of Ba²⁺ ions and 2 moles of Br⁻ ions.

So we can say that:

moles of Ba²⁺ = moles of BaBr₂

moles of Ba²⁺ = 8.01 * 10⁻⁷ moles

moles of Br⁻ = 2 * moles of BaBr₂

moles of Br⁻ = 2 * 8.01 * 10⁻⁷ moles

moles of Br⁻ = 1.60 * 10⁻⁶ moles

Now we can find the molarity of each ion:

Molarity of Ba²⁺ = moles of Ba²⁺ / volume of solution in L

Molarity of Ba²⁺ = 8.01 * 10⁻⁷ moles/(2.00 L)

Molarity of Ba²⁺ = 4.00 * 10⁻⁷ M

Molarity of Br⁻ = moles of Br⁻ / volume of solution in L

Molarity of Br⁻ = 1.60 * 10⁻⁶ moles / (2.00 L)

Molarity of Br⁻ = 8.00 * 10⁻⁷ M

Answer: The concentration of BaBr₂ and the concentration of Ba²⁺ is 4.00 * 10⁻⁷ M. The concentration of Br⁻ is 8.00 * 10⁻⁷ M.

When we express the concentration in ppm we want the concentration in mg/L. We have the concentration in M (moles/L), so to find the concentration in ppm we have to convert moles to mg.

The molar masses of the ions are:

molar mass of Ba²⁺ = 137.33 g/mol

molar mass of Br⁻ = 79.90 g/mol

Let's use them to find the concentration in ppm.

concentration of Ba²⁺ = 4.00 * 10⁻⁷ M = 4.00 * 10⁻⁷ moles/L * 137.33 g/mol

concentration of Ba²⁺ = 5.49 * 10⁻⁵ g/L *1000 mg/g

concentration of Ba²⁺ = 0.0549 mg/L = 0.0549 ppm

concentration of Ba²⁺ = 0.0549 ppm

concentration of Br⁻ = 8.00 * 10⁻⁷ M = 8.00 * 10⁻⁷ moles/L * 79.90 g/mol

concentration of Br⁻ = 6.39 * 10⁻⁵ g/L * 1000 mg/g

concentration of Br⁻ = 0.0639 mg/L = 0.0639 ppm

concentration of Br⁻ = 0.0639 ppm

Answer: the concentration of the ion Ba²⁺ is 0.0549 ppm and the concentration of the ion Br⁻ is 0.0639 ppm.