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2 years ago

solubility of cu(oh)2 is 2.42 x 10-3 mol/l. calculate ksp and in which direction the reaction is spontaneous if δgº

Chemistry

1 answer:

Simora [160]2 years ago

7 0

Ksp of copper(II) hydroxide Cu(OH)2 is 4.9 x 10-8.

Chemical reaction (dissociation) of copper(II) hydroxide in water:

Cu(OH)2(s) → Cu²⁺(aq) + 2OH⁻(aq).

Ksp(Cu(OH)2) = [Cu²⁺]·[OH⁻]².

[Cu²⁺] = 2.42 x 10-3 mol/l; solubility od copper ions

[OH⁻] = 2[Cu²⁺] = 2 x 2.42 x 10-3 mol/l

[OH⁻] = 4.48 x 10-3 mol/l; solubility of hydroxide ions

Ksp = 2.42 x 10-3 mol/l x (4.48 x 10-3 mol/l)²

Ksp = 4.9 x 10-8

Ksp is the solubility product constant for a solid substance dissolving in an aqueous solution.

Solubility of the compound depends on the temperature of the solution and the structure of that compound.

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What is the molarity of the Ca(OH)2 solution if 32.00 mL of Ca(OH)2 requires 16.08 mL of a 2.303 M solution for complete titrati

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Answer:

For this problem we just need to remember the equation and that the volume is always in liters: MaVa=MbVb

Ma= 1.338 mol/L Va= 18.75 mL= 0.01875 L Mb= x Vb=24.73 mL= 0.02473 L

So now we can plug into the equation and solve:

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This is a two step process: stoichiometry and using the answer from the first part to plug and chug it into the Molariy equation.

First step: setup the stoichiometry problem:

3.1171 g Na2CO3 * (1 mol/106 g) * (2 mol HCl/ 1 mol Na2CO3)= mol HCl

Second step: Molarity equation

Molarity= moles HCl/ L M= mol HCl/0.04027 L

For the third problem, you will just use the same equation as the first: MaVa= MbVb

Ma=0.57 M Va= x L Mb= 0.875 M Vb= 23.83 mL= 0.02383 L

0.57 M * x L= 0.875 M * 0.02383 L

With this equation, we want to find the moles of NaOH by using the molarity equation first then because MaVa= MbVb, we know that the number of moles has to be equal.

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So next, we can set it to the mass of the acid using this equation: Molar Mass= mass/ moles

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And with that you will find the molar mass of HX, and even determine what X is.

Explanation:

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Answer:

a. 581.4 Pa

b. 3.33x10⁻⁴ mol/L

c. 3.49x10⁻⁴ mol/L

d. 0.015 g/L

Explanation:

a. By the Raoult's Law, the partial pressure of a component of a gas mixture is its composition multiplied by the total pressure, so:

pA = 0.9532*6.1

pA = 5.81452 mbar

pA = 5.814x10⁻³ bar

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5.814x10⁻³ bar--- pA

pA = 581.4 Pa

b. Considering the mixture as an ideal gas, let's assume the volume as 1,000 L, so by the ideal gas law, the total number of moles is:

PV = nRT

Where P is the pressure (610 Pa), V is the volume (1 m³), n is the number of moles, R is the gas constant (8.314 m³.Pa/mol.K), and T is the temperature.

n = PV/RT

n = (610*1)/(8.314*210)

n = 0.3494 mol

The number of moles of CO₂ is (V = 0.9532*1 = 0.9532 m³):

n = PV/RT

n = (581.4*0.9532)/(8.314*210)

n = 0.3174 mol

cA = n/V

cA = 0.3174/953.2

cA = 3.33x10⁻⁴ mol/L

c. c = ntotal/Vtotal

c = 0.3494/1000

c = 3.49x10⁻⁴ mol/L

d. The molar masses of the gases are:

CO₂: 44 g/mol

N₂: 28 g/mol

Ar: 40 g/mol

O₂: 32 g/mol

CO: 28 g/mol

The molar mass of the mixture is:

M = 0.9532*44 + 0.027*28 + 0.016*40 + 0.0008*28 = 43.36 g/mol

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