Calculate the Standard Enthalpy of the reaction below:

Answer the following questions about the solubility of AgCl(s). The value of Ksp for AgCl(s) is 1.8 × 10−10.

Firlakuza [10]

Answer:

[Ag⁺] = 1.3 × 10⁻⁵M

s = 3.3 × 10⁻¹⁰ M

Because the common ion effect.

Explanation:

1. Value of [Ag⁺] in a saturated solution of AgCl in distilled water.

The value of [Ag⁺] in a saturated solution of AgCl in distilled water is calculated by the dissolution reaction:

AgCl(s) ⇄ Ag⁺ (aq) + Cl⁻ (aq)

The ICE (initial, change, equilibrium) table is:

AgCl(s) ⇄ Ag⁺ (aq) + Cl⁻ (aq)

I X 0 0

C -s +s +s

E X - s s s

Since s is very small, X - s is practically equal to X and is a constant, due to which the concentration of the solids do not appear in the Ksp equation.

Thus, the Ksp equation is:

Ksp = [Ag⁺] [Cl⁻]

Ksp = s × s

Ksp = s²

By substitution:

1.8 × 10⁻¹⁰ = s²

s = 1.34 × 10⁻⁵M

Rounding to two significant figures:

[Ag⁺] = 1.3 × 10⁻⁵M ← answer

2. Molar solubility of AgCl(s) in seawater

Since, the conentration of Cl⁻ in seawater is 0.54 M you must introduce this as the initial concentration in the ECE table.

The new ICE table will be:

AgCl(s) ⇄ Ag⁺ (aq) + Cl⁻ (aq)

I X 0 0.54

C -s +s +s

E X - s s s + 0.54

The new equation for the Ksp equation will be:

Ksp = [Ag⁺] [Cl⁻]

Ksp = s × ( s + 0.54)

Ksp = s² + 0.54s

By substitution:

1.8 × 10⁻¹⁰ = s² + 0.54s

s² + 0.54s - 1.8 × 10⁻¹⁰ = 0

Now you must solve a quadratic equation.

Use the quadratic formula:

$s=\dfrac{-0.54\pm\sqrt{0.54^2-4(1)(-1.8\times 10^{-10})}}{2(1)}$

The positive and valid solution is s = 3.3×10⁻¹⁰ M ← answer

3. Why is AgCl(s) less soluble in seawater than in distilled water.

AgCl(s) is less soluble in seawater than in distilled water because there are some Cl⁻ ions is seawater which shift the equilibrium to the left.

This is known as the common ion effect.

By LeChatelier's principle, you know that an increase in the concentrations of one of the substances that participate in the equilibrium displaces the reaction to the direction that minimizes this efect.

In the case of solubility reactions, this is known as the common ion effect: when the solution contains one of the ions that is formed by the solid reactant, the reaction will proceed in less proportion, i.e. less reactant can be dissolved.

3 0

2 years ago

Building Vocabulary

Juliette [100K]

Answer:

Explanation:

Building Vocabulary

Match each term with its definition by writing the letter of the correct definition on

the line beside the term in the left column.

5. nucleus b

6. proton f

7. neutron h

8. electron d

9. atomic number g

10. isotopes c

11. mass number a

12. energy level e

a. the sum of protons and neutrons in the nucleus of an

atom

b. the very small center core of an atom

c. atoms of the same element that differ in the number

of neutrons, but have the same number of protons

d. the particle of an atom that moves rapidly in the

space outside the nucleus

e. a specific amount of energy related to the movement

of electrons in atoms

f. the particle of an atom with a positive charge

g. the number of protons in the nucleus of every atom

of an element

h. the particle of an atom that is neutral

-. mass number a.

12. energy level e

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What is the percent by mass of sodium in Na2SO4? total mass of element in compound molar mass of compound Use %Element x 100

Kisachek [45]

What is the percent by mass of sodium in Na2SO4? total mass of element in compound molar mass of compound Use %Element x 100

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Reema took 5ml of Lead Nitrate solution in a beaker and added approximately 4ml of Potassium Iodide solution to it. What would s

Brrunno [24]

She would observe a yellowish solid precipitate which is the lead iodide and a white solid precipitate which is the potassium nitrate.

This is because the lead nitrate solution which contains particles of lead will mix with the potassium iodide solution containing particles of iodide. Upon mixing,the lead particles from the Lead nitrate solution combines with the iodide particles from Potassium iodide and form two compounds, a yellowish solid precipitate called lead iodide and a white solid precipitate called Potassium nitrate.

The formation of entirely two new compounds is known as the double displacement reaction and can be written in a chemical equation as

2KI(aq.)+Pb(NO₃)₂(aq.)------>2KNO₃(aq.)+PbI ₂(s)

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