Question

1. (a) What is the Na⁺ concentration in each of the following solutions:

2.85 M sodium sulfate:

2.18 M sodium carbonate:

0.985 M sodium bicarbonate:

(b) What is the concentration of a lithium carbonate solution that is 0.395 M in Li⁺? M

2. A sample of 0.7160 g of an unknown compound containing barium ions (Ba²⁺) is dissolved in water and treated with an excess of Na₂SO₄. If the mass of the BaSO₄ precipitate formed is 0.6894 g, what is the percent by mass of Ba in the original unknown compound?

3. A 0.9080 g sample of a mixture of NaCl and KCl is dissolved in water, and the solution is then treated with an excess of AgNO₃ to yield 1.953 g of AgCl. Calculate the percent by mass of each compound in the mixture.

% mass NaCl

% mass KCl

4. The concentration of lead ions (Pb²⁺) in a sample of polluted water that also contains nitrate ions (NO₃⁻) is determined by adding solid sodium sulfate (Na₂SO₄) to exactly 500 mL of the water. Calculate the molar concentration of Pb²⁺ if 0.00260 g of Na₂SO₄ was needed for the complete precipitation of lead ions as PbSO₄.

5. Consider the reaction

H₂(g) + Cl₂(g) → 2HCl(g) Δ H = −184.6 kJ/mol

If 5.0 moles of H₂ reacts with 5.0 moles of Cl₂ to form HCl at 1.0 atm, what is Δ U for this reaction? Assume the reaction goes to completion and Δ V = 0 L.
(The conversion factor is 1 L atm = 101.3 J.)

Answer 1

1) Sodium concentrations are calculated:

a) Na2SO4:

[Na +] = 2 * 2.85 = 5.7 M

Na2CO3:

[Na +] = 2 * 2.18 = 4.36 M

NaHCO3:

[Na +] = 0.985 M

b) The concentration of lithium carbonate is calculated:

[Li2CO3] = 0.395 / 2 = 0.1975 M

2) The mass of Ba of the precipitate is calculated:

m Ba = 0.6894 g * (137.3 g Ba / 233.4 g BaSO4) = 0.4055 g

The percentage is calculated:

% Ba = 0.4055 * 100 / 0.7160 = 56.63%

3) The mass of Cl is calculated:

m Cl = 1,953 g AgCl * (35.45 g Cl / 143.32 g AgCl) = 0.4831 g Cl

It has to:

i) m NaCl + m KCl = 0.9080

ii) (MM Cl / MM NaCl) * m NaCl + (MM Cl / MM KCl) * m KCl = m Cl

0.61 * m NaCl + 0.48 * m KCl = 0.4831

System of equations between 1 and 2 is applied and you have:

m NaCl = 0.3635 g

m KCl = 0.5445 g

The percentage is calculated:

% NaCl = 0.3635 * 100 / 0.9080 = 40%

% KCl = 100 - 40 = 60%

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