At 25 oC the solubility of copper(I) chloride is 1.00 x 10-3 mol/L. Calculate the value of Ksp at this temperature. Give your an
1 answer:
You might be interested in
Answer:
1.82 g is the maximum mass of CuS.
Explanation:
Considering:
Or,
Given :
<u>For :
</u>
Molarity = 0.500 M
Volume = 38.0 mL
The conversion of mL to L is shown below:
1 mL = 10⁻³ L
Thus, volume = 38.0×10⁻³ L
Thus, moles of :
<u>Moles of = 0.019 moles
</u>
<u>For :
</u>
Molarity = 0.600 M
Volume = 42.0 mL
The conversion of mL to L is shown below:
1 mL = 10⁻³ L
Thus, volume = 42.0×10⁻³ L
Thus, moles of :
<u>Moles of = 0.0252 moles
</u>
According to the given reaction:
1 mole of reacts with 1 mole of
So,
0.019 mole of reacts with 0.019 mole of
Moles of = 0.019 mole
Available moles of = 0.0252 mole
<u>Limiting reagent is the one which is present in small amount. Thus, is limiting reagent.</u>
The formation of the product is governed by the limiting reagent. So,
1 mole of gives 1 mole of
0.019 mole of gives 0.019 mole of
Moles of formed = 0.019 moles
Molar mass of = 95.611 g/mol
Mass of = Moles × Molar mass = 0.019 × 95.611 g = 1.82 g
<u>1.82 g is the maximum mass of CuS.</u>
The equilibrium constant is 0.0022.
Explanation:
The values given in the problem is
ΔG° = 1.22 ×10⁵ J/mol
T = 2400 K.
R = 8.314 J mol⁻¹ K⁻¹
The Gibbs free energy should be minimum for a spontaneous reaction and equilibrium state of any reaction is spontaneous reaction. So on simplification, the thermodynamic properties of the equilibrium constant can be obtained as related to Gibbs free energy change at constant temperature.
The relation between Gibbs free energy change with equilibrium constant is ΔG° = -RT ln K
So, here K is the equilibrium constant. Now, substitute all the given values in the corresponding parameters of the above equation.
We get,
So, the equilibrium constant is 0.0022.