Question

6 b)
In the laboratory a student combines 46.6 mL of a 0.322 M potassium sulfate solution with 19.0 mL of a 0.503 M cobalt(II) sulfate solution.

What is the final concentration of sulfate anion ?

M

18 a) A 0.5655 g sample of a pure soluble bromide compound is dissolved in water, and all of the bromide ion is precipitated as AgBr by the addition of an excess of silver nitrate. The mass of the resulting AgBr is found to be 0.9510 g.

What is the mass percentage of bromine in the original compound? %

18 b) A student determines the nickel content of a solution by first precipitating it as nickel hydroxide, and then decomposing the hydroxide to nickel oxide by heating. How many grams of nickel oxide should the student obtain if her solution contains 48.0 mL of 0.587 M nickel nitrate?

g

Answer 1

6b) Consider the ionizations of potassium sulfate, K₂SO₄ and cobalt (II) sulfate, CoSO₄ as below.

K₂SO₄ (aq) ---------> 2 K⁺ (aq) + SO₄^2- (aq)

CoSO₄ (aq) ---------> Co²⁺ (aq) + SO₄^2- (aq)

As per the stoichiometric equations,

1 mole K₂SO₄ = 1 mole SO₄^2-.

Again,

1 mole CoSO₄ = 1 mole SO₄^2-.

Millimol(s) K₂SO₄ = millimol(s) SO₄^2- = (46.6 mL)*(0.322 M) = 15.0052 mmol.

Millimol(s) CoSO₄ = millimole(s) SO₄^2- = (19.0 mL)*(0.503 M) = 9.557 mmol.

Total millimol(s) SO₄^2- = (15.0052 + 9.557) mmol = 24.5622 mmol.

Total volume of the solution = (46.6 + 19.0) mL = 65.6 mL.

Final concentration of SO₄^2- in the solution = (24.5622 mmol)/(65.6 mL)

= 0.3744 M ≈ 0.374 M (ans, correct to 3 sig. figs).

18a) Mass of sample = 0.5655 g.

Mass of silver bromide, AgBr precipitated = 0.9510 g.

The atomic masses are

Ag: 107.868 u

Br: 79.904 u

The gram molar mass of AgBr = (1*107.868 + 1*79.904) g/mol = 187.772 g/mol.

187.772 g AgBr contains 79.904 g bromide.

Therefore, mass of bromide in 0.9510 g AgBr = (0.9510 g)*(79.904 g)/(187.772 g)

= 0.4047 g.

Since there are no other sources of bromine, hence, the entire bromine comes from the sample. Therefore, 0.5655 g sample contains 0.4047 g bromine (as bromide).

Therefore, mass percentage of bromine in the sample = (0.4047 g)/(0.5655 g)*100

= 71.565% ≈ 71.56% (ans, correct to 4 sig. figs).

18b) The sequence of reactions is shown below.

Ni(NO₃)₂ (aq) + 2 OH^- (aq) --------> Ni(OH)₂ (s) + 2 NO₃^- (aq) ……(1)

Ni(OH)₂ (s) --------> NiO (s) + H₂O (l) …….(2)

As per the stoichiometric equations above,

1 mol Ni(NO₃)₂ = 1 mol Ni(OH)₂ = 1 mol NiO.

Millimol(s) Ni(NO₃)₂ = (48.0 mL)*(0.587 M) = 28.176 mmol.

Therefore, millimole(s) NiO formed = 28.176 mmol.

The atomic masses are

Ni: 58.693 u

O: 15.999 u’

The gram molar mass of NiO = (1*58.693 + 1*15.999) g/mol = 74.692 g/mol.

Mass of NiO obtained from 48.0 mL of 0.587 M Ni(NO₃)₂ = (28.176 mmol)*(74.692 g/mol)

= (28.176 mmol)*(1 mol)/(1000 mmol)*(74.692 g/mol)

= 2.1045 g ≈ 2.10 g (ans, correct to 3 sig. figs).