Question

Excess Na2SO4 (aq) is added to a 42.53 mL sample of Ba(NO3)2(aq).

1. What is the formula of the precipitate?
The formula for the precipitate is 2Na(NO3)(aq).

2. If 3.046 grams of precipitate was formed, what was the molarity of the Ba(NO3)2(aq)?

Answer 1

1. The formula of the precipitate formed is Ba₂SO₄(s)

2. The molarity of the Ba(NO₃)₂ solution is 0.188 M

1. Determination of the formula of the precipitate formed.

• We'll begin by writing the net ionic equation for the reaction between Na₂SO₄ and Ba(NO₃)₂. This is given below:

Na₂SO₄(aq) + Ba(NO₃)₂(aq) —>

2Na⁺(aq) + SO₄²¯(aq) + Ba²⁺(aq) + 2NO₃¯(aq) —> Ba₂SO₄(s) + 2Na⁺(aq) + 2NO₃¯(aq)

Cancel the spectator ions (i.e Na⁺ and NO₃¯) to obtain the net ionic equation

SO₄²¯(aq) + Ba²⁺(aq) —> Ba₂SO₄(s)

Therefore, the formula of the precipitate formed is Ba₂SO₄(s)

2. Determination of the molarity of Ba(NO₃)₂

• We'll begin by calculating the number of mole of in 3.046 g of precipitate, Ba₂SO₄

Molar mass of Ba₂SO₄ = (2×137) + 32 + (4×16) = 370 g/mol

Mass of Ba₂SO₄ = 3.046 g

Mole of Ba₂SO₄ =?

Mole = mass /molar mass

Mole of Ba₂SO₄ = 3.046 / 370

Mole of Ba₂SO₄ = 0.008 mole

• Next, we shall determine the number of mole of Ba(NO₃)₂ required to produce 0.008 mole of Ba₂SO₄

Na₂SO₄(aq) + Ba(NO₃)₂(aq) —> Ba₂SO₄(s) + 2NaNO₃(aq)

From the balanced equation above,

1 mole of Ba(NO₃)₂ reacted to produce 1 mole Ba₂SO₄.

Therefore,

0.008 mole of Ba(NO₃)₂ will also react to produce 0.008 mole of Ba₂SO₄

Thus, 0.008 mole of Ba(NO₃)₂ is required for the reaction.

• Finally, we shall determine the molarity of the Ba(NO₃)₂ solution

Mole of Ba(NO₃)₂ = 0.008 mole

Volume = 42.53 mL = 42.53 / 1000 = 0.04253 L

Molarity of Ba(NO₃)₂ =?

Molarity = mole / Volume

Molarity of Ba(NO₃)₂ = 0.008 / 0.04253

Molarity of Ba(NO₃)₂ = 0.188 M

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