Equilibrium Test 9

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Equilibrium Test 9

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Weekly Quiz Competition

Question 1
1 / -0

. What is the molar solubility (s) of Ba₃(PO₄)₂ in terms of K_sp?

A
s = [K_sp/27]^1/5

B
s = K_sp^1/5

C
s = K_sp^1/2

D
s = [K_sp/108]^1/5

Solution

Ba₃(PO₄)₂ ⇌ 3Ba⁺² (2PO₄)^-3

K_sp = [Ba⁺²>]³ [(PO₄)^-3]²
[3s]³[2s]²
K_sp = 108s⁵
Or s = (K_sp/108)^⅕
Question 2
1 / -0

What is the K_sp expression for the salt PbI₂?

A
[Pb²⁺][I^–]²

B
[Pb²⁺][I₂]²

C
[Pb²⁺][2I^–]

D
[Pb²⁺][2I^–]²

Solution

The correct answer is Option D.
PbI2 ‡Œ Pb²⁺ + 2I^–
or,
Ksp = [Pb²⁺][2I^–]²
Question 3
1 / -0

The solubility product expression for tin(II) hydroxide, Sn(OH)₂, is

A
[Sn²⁺][OH-]

B
[Sn²⁺][OH^-]²

C
[Sn²⁺]³[OH^–]

D
[Sn²⁺]²[OH^–]

Solution
Question 4
1 / -0

The [Ag⁺_(aq)] = 10^-5 in a solution .The [Cl^–_(aq)] to precipitate AgCl having K_sp of 1.8×10¹⁰ M² is — M

A
10^-7

B
10^-8

C
10^-9

D
10^-5

Solution

For precipitation, Qrkn > Ksp, then tto establish equilibrium, ions will combine to give molecule as ppt.
So applying the above concept,
1.8×10^-10 < [Ag⁺][Cl^-]
1.8×10^-10 < 10^-5 × [Cl^-]
Or [Cl-] > 1.8×10^-5
By seeing the option, only option d satisfies this condtion.
Question 5
1 / -0

The molar solubility of PbBr₂ is 2.17 x 10^-3 M at a certain temperature. Calculate K_spfor PbBr₂

A
4.1 x 10^-8

B
6.2 x 10^-6

C
6.4 x 10^-7

D
3.4 x 10⁶

Solution

Molar solubility(s) for PbBr₂ = 2.17×10^-3 M
PbBr₂ ⇌ Pb⁺²+ 2Br^-
ss 2s
K_sp = [Pb⁺²][Br^-]²
= [s][s]²
= 4s³ = 4(2.17×10^-3)³
= 4.1×10^-8
Question 6
1 / -0

What is the correct equilibrium expression (K_sp) for the reaction below:
Ca₃(PO₄)₂(s) 3Ca²⁺(aq) + 2PO₄^2-(aq)

A
K_sp = [Ca²⁺]³[PO₄^3-]²

B
K_sp = [3Ca²⁺] + [2PO₄^3-]

C
K_sp = [3Ca²⁺][2PO₄^3-]

D
K_sp = [3Ca²⁺]³ + [2PO₄^3-]²
Question 7
1 / -0

What is the Ksp expression for lead (II) chloride?

A
[Pb²⁺][Cl^–]²

B
[Pb²⁺]2[Cl^–]

C
[Pb+][Cl-]²/[PbCl₂]

D
[Pb2²⁺][Cl^–]²

Solution

PbCl₂⇌ Pb²⁺+ 2Cl^–or, K_sp = [Pb²⁺][Cl^–]²
Question 8
1 / -0

Solubility of BaCl₂ if Ksp is 10^-6 at 25°C is

A
Cannot be predicted

B
6.3 10^-3 M

C
10^-6 M

D
10^-3 M

Solution

The correct answer is Option B.
BaCl₂ ⇔ Ba²⁺ + 2Cl⁻
Ksp = (s)(2s)² = 4s³
Given, 4s³ = 10⁻⁶
⇒s³ = 0.25 ⨯ 10⁻⁶
⇒s = (0.25)³ ⨯ 10⁻²
⇒s = 0.629 ⨯ 10⁻²
⇒s = 6.3 ⨯ 10⁻³
Question 9
1 / -0

When in a saturated solution of NaCl, HCl is passed, pure precipitate of NaCl is formed. This is due to the fact:

A
That solubility of NaCl decreases

B
That HCl is strong acid

C
That the ionic product of NaCl exceeds the solubility product of NaCl.

D
That HCl is highly soluble in water

Solution

In saturated solution of NaCl,equillibrium is established between ions and unionised NaCl. When HCl gas is passed through saturated solution of NaCl,it is completely dissociated to form H+ and Cl- ions. Cl- ion is common to both electrolyte ,hence Cl- ion concentration in solution increases.
Question 10
1 / -0

Solubility product is defined as

A
product of molar concentration of ions of a salt as per stoichiometry in a supersaturated solution

B
product of molar concentration of ions of a salt as per stoichiometry in a saturated solution at STP

C
product of molar concentration of ions of a salt as per stoichiometry in a saturated solution

D
product of molar concentration of ions of a salt as per stoichiometry in an unsaturated solution

Solution

Definition of Solubility Product, K.A substance's solubility product, Ksp, is the mathematical product of its dissolved ion concentrations raised to the power of their stoichiometric coefficients. Solubility products are relevant when a sparingly soluble ionic compound releases ions into solution.
Question 11
1 / -0

The solubility product expression for silver(I) sulphide, using x to represent the molar concentration of silver(I) and y to represent the molar concentration of sulphide, is formulated as:

A
x²y²

B
xy³

C
x²y

D
xy²

Solution

The correct answer is Option C.
The ionization equilibrium of silver (I) sulfide is
Ag₂S⇌2Ag⁺ +S2⁻
The solubility product expression for silver (I) sulfide is KSP[Ag⁺]2 [S2⁻].
But [Ag⁺]² = x and [S²⁻] = y.
Hence the expression for the solubility product becomes K_SP[Ag⁺]² [S²⁻] = x²y.
Question 12
1 / -0

Consider the following solubility data for various chromates at 25°C.

A
Ag₂CrO₄

B
BaCrO₄

C
PbCrO₄

D
none of these

Solution

The salt with more K_sp is more soluble in water. Hence BaCrO₄ is more soluble.
Question 13
1 / -0

A solution is 0.0010 M in both Ag⁺ and Au⁺. Some solid NaCl is added slowly until the solid AgCl just begins to precipitate. What is the concentration of Au⁺ ions at this point? K_sp for AgCl = 1.8 x 10^-10 and for AuCl = 2.0 x 10^-13.

A
2.0 x 10^-10 M

B
4.5 x 10^-7 M

C
1.8 x 10^-7 M

D
1.1 x 10^-6 M

Solution
Question 14
1 / -0

Precipitation requires

A
Ionic product to be less than solubility product

B
Ionic product to be equal to solubility product

C
Ionic product to be more than pH

D
Ionic product to be more than solubility product

Solution

PRECIPITATION: Precipitation occurs when tiny droplets of water, ice, or frozen water vapor join together into masses too big to be held above the earth. They then fall to the ground as precipitation.
Question 15
1 / -0

Solubility of a soluble salt is

A
more than 0.1M

B
less than 0.01M

C
more than 0.001M

D
more than 0.01M

Solution

A salt is soluble if it dissolves in water to give a solution with a concentration of at least 0.1 moles per liter at room temperature.

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Equilibrium Test 9

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Equilibrium Test 9

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Question 1

s = [Ksp/27]1/5

s = Ksp1/5

s = Ksp1/2

s = [Ksp/108]1/5

Solution

Question 2

[Pb2+][I–]2

[Pb2+][I2]2

[Pb2+][2I–]

[Pb2+][2I–]2

Solution

Question 3

[Sn2+][OH-]

[Sn2+][OH-]2

[Sn2+]3[OH–]

[Sn2+]2[OH–]

Solution

Question 4

10-7

10-8

10-9

10-5

Solution

Question 5

4.1 x 10-8

6.2 x 10-6

6.4 x 10-7

3.4 x 106

Solution

Question 6

Ksp = [Ca2+]3[PO43-]2

Ksp = [3Ca2+] + [2PO43-]

Ksp = [3Ca2+][2PO43-]

Ksp = [3Ca2+]3 + [2PO43-]2

Question 7

[Pb2+][Cl–]2

[Pb2+]2[Cl–]

[Pb+][Cl-]2/[PbCl2]

[Pb22+][Cl–]2

Solution

Question 8

Cannot be predicted

6.3 10-3 M

10-6 M

10-3 M

Solution

Question 9

That solubility of NaCl decreases

That HCl is strong acid

That the ionic product of NaCl exceeds the solubility product of NaCl.

That HCl is highly soluble in water

Solution

Question 10

product of molar concentration of ions of a salt as per stoichiometry in a supersaturated solution

product of molar concentration of ions of a salt as per stoichiometry in a saturated solution at STP

product of molar concentration of ions of a salt as per stoichiometry in a saturated solution

product of molar concentration of ions of a salt as per stoichiometry in an unsaturated solution

Solution

Question 11

x2y2

xy3

x2y

xy2

Solution

Question 12

Ag2CrO4

BaCrO4

PbCrO4

none of these

Solution

Question 13

s = [K_sp/27]^1/5

s = K_sp^1/5

s = K_sp^1/2

s = [K_sp/108]^1/5

[Pb²⁺][I^–]²

[Pb²⁺][I₂]²

[Pb²⁺][2I^–]

[Pb²⁺][2I^–]²

[Sn²⁺][OH-]

[Sn²⁺][OH^-]²

[Sn²⁺]³[OH^–]

[Sn²⁺]²[OH^–]

10^-7

10^-8

10^-9

10^-5

4.1 x 10^-8

6.2 x 10^-6

6.4 x 10^-7

3.4 x 10⁶

K_sp = [Ca²⁺]³[PO₄^3-]²

K_sp = [3Ca²⁺] + [2PO₄^3-]

K_sp = [3Ca²⁺][2PO₄^3-]

K_sp = [3Ca²⁺]³ + [2PO₄^3-]²

[Pb²⁺][Cl^–]²

[Pb²⁺]2[Cl^–]

[Pb+][Cl-]²/[PbCl₂]

[Pb2²⁺][Cl^–]²

6.3 10^-3 M

10^-6 M

10^-3 M

x²y²

xy³

x²y

xy²

Ag₂CrO₄

BaCrO₄

PbCrO₄

2.0 x 10^-10 M

4.5 x 10^-7 M

1.8 x 10^-7 M

1.1 x 10^-6 M