The solubility of KCl is 3.7 M at 20 °C. Two beakers each contain 100. mL of saturated KCl solution: 100. mL of 4.0 M HCl is add
2 answers:
Answer:
(a) 13.69
(b) i beaker 1: 0g
ii beaker 2: 0g
Explanation:
a. The KCl solubility equilibrium equation is
KCl(s) ⇄ K⁺(aq) + Cl⁻(aq)
A 3.7M KCl solution contains equal amounts of K ions and Cl ions hence
the ion-product formula can be outlined as
Ksp = [K⁺][Cl⁻]
= [3.7][3.7]
= 13.69
b. From the two beakers, each holding 100 mL and 3.7M KCl
the molarity of K⁺ = molarity of Cl⁻ = moles of KCl = 3.7moles in 1L,
noting that 3.7M represents 3.7 mol in 1L or 1000mL or 1000 cm³.
Calculating the total moles contained in 100 mL:
3.7moles/Liter * 100 mL
= 0.37moles K⁺ = 0.37moles Cl⁻.
A 4.0 M HCl solution contains
in 100mL.
An 8.0M HCl solution contains
in 100mL.
In the first beaker, 100 mL of 4M HCl is combined with 100 mL of 3.7M KCl:
Total moles of Cl⁻ (0.4 + 0.37) equals 0.77 moles.
While total moles of K⁺ remain 0.37 moles.
The total solution volume = (100mL + 100mL) = 200mL/1000mL = 0.2L.
Thus moles of Cl⁻ calculated per liter = 0.77moles/0.2L = 3.85M Cl⁻.
Moles of K⁺ calculated per liter = 0.37moles/0.2L = 1.85M K⁺.
For precipitation to take place, Qsp should be equal to or exceed Ksp, which means:
Qsp ≥ Ksp
Qsp = [K][Cl] = [1.85][3.85] = 7.12, which remains below 13.69 (Ksp).
Consequently, no KCl will precipitate in the first beaker, and since there’s no precipitate, calculating the mass that would have formed is unnecessary.
Thus, the result is 0g.
(bii) In the second beaker, 100 mL of 8M HCl is combined with 100 mL of 3.7M KCl:
Total moles of Cl⁻ (0.8 + 0.37) total to 1.17 moles.
With total moles of K⁺ remaining at 0.37 moles.
Providing the total solution volume = (100mL + 100mL) = 200mL/1000mL = 0.2L.
Total moles of Cl⁻ per liter = 1.17moles/0.2L = 5.85M Cl⁻.
Total moles of K⁺ per liter = 0.37moles/0.2L = 1.85M K⁺.
For precipitation to happen, Qsp needs to be at minimum equal to Ksp, which implies:
Qsp ≥ Ksp
Qsp = [K][Cl] = [1.85][5.85] = 10.82 which, again, is below 13.69 (Ksp).
Thus, no KCl will precipitate in the second beaker; thus, calculating mass for precipitation is also rendered unnecessary.
The answer here is also 0g.
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The two states possess distinct densities, thus occupying varying volumes; however, the mass remains constant.
Step 1: Calculate the mass of nitrogen gas
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Using these values yields:
The mass of nitrogen gas calculates to be 1356 g.
Step 2: Derive the volume of liquid nitrogen from the mass obtained
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With the density of liquid nitrogen at 809 g/L, we can substitute this into our formula to find the volume of liquid.
Therefore, the volume of liquid nitrogen is 1.68 L.