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Test: Ionic Equilibrium - 2 - Chemistry MCQ


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20 Questions MCQ Test Topicwise Question Bank for IIT JAM/CSIR/GATE Chemistry - Test: Ionic Equilibrium - 2

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Test: Ionic Equilibrium - 2 - Question 1

What percentage of ammonia is given after cooling equilibrium mixture by Haber process?

Detailed Solution for Test: Ionic Equilibrium - 2 - Question 1

The Haber Process combines nitrogen from the air with hydrogen derived mainly from natural gas (methane) into ammonia. The reaction is reversible and the production of ammonia is exothermic.
At each pass of the gases through the reactor, only about 15% of the nitrogen and hydrogen converts to ammonia. By continual recycling of the unreacted nitrogen and hydrogen, the overall conversion is about 98%.

Test: Ionic Equilibrium - 2 - Question 2

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

Detailed Solution for Test: Ionic Equilibrium - 2 - Question 2

Ba3(PO4)2 ⇌ 3Ba+2  (2PO4)-3

Ksp =  [Ba+2>]3 [(PO4)-3]2
[3s]3[2s]2
Ksp = 108s5
Or s = (Ksp/108)

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Test: Ionic Equilibrium - 2 - Question 3

 What is the Ksp expression for the salt PbI2?

Detailed Solution for Test: Ionic Equilibrium - 2 - Question 3

A general equation of equilibrium constant for a reaction of dissociation of PbI2 as PbI2⇌Pb2++2I-can be written as,

But we know that Ksp is the equilibrium constant at the saturated level. At saturation, no more PbIwill dissolve. Thus the concentration of PbI2 will be constant and can be taken as 1.
Therefore the above equation can be written as,

Test: Ionic Equilibrium - 2 - Question 4

Which of the following is more acidic? A solution with pH 5 or a solution with pH 3

Detailed Solution for Test: Ionic Equilibrium - 2 - Question 4

We know that pH equals to - log[H+] or the value of pH is inversely proportional to the concentration of H+.  So more the value of pH, less the solution be acidic.

Test: Ionic Equilibrium - 2 - Question 5

A 0.10 M solution of a weak acid, HX, is 0.059% ionized. Evaluate Ka for the acid.

Detailed Solution for Test: Ionic Equilibrium - 2 - Question 5

Since the acid is only 0.059% ionized, therefore the concentration of ions in solution = 0.1 x 0.059 / 100 = 0.000059

 

Ka = [H+] [X] / [HX] = (0.000059)2 / 0.1 = 3.5 x 10-8

Test: Ionic Equilibrium - 2 - Question 6

 In a sample of pure water, only one of the following statements is always true at all conditions of temperature and pressure.

Detailed Solution for Test: Ionic Equilibrium - 2 - Question 6

The correct answer is Option D.
The concentration of OH- will always equal the concentration of H3O+.
PV = nRT, where R is a constant, and n is the quantity of gas measured in mols. The pH of a substance is related to the concentration of H+,
so if we raise or lower the Temperature or Pressure of pure water, we will alter the Volume (or concentration) of H3O+ and OH- ions. As we alter the concentrations we alter the pH and pOH. The only answer that is true no matter what the volume or concentration may be is D
 

Test: Ionic Equilibrium - 2 - Question 7

The pH of 10-8 molar solution of HCl in water is:

Detailed Solution for Test: Ionic Equilibrium - 2 - Question 7

If we use the relation, pH = – log [H3O+], we get pH equal to 8. But this is not correct because an acidic solution cannot have pH greater than 7. It may be noted that in very dilute acidic solution, when H+ concentrations from acid and water are comparable, the concentration of H+ from water cannot be neglected.

Therefore,

Test: Ionic Equilibrium - 2 - Question 8

The pH of a 0.02 M solution of an unknown weak acid is 3.7. What is the pKa of this acid?

Detailed Solution for Test: Ionic Equilibrium - 2 - Question 8

Test: Ionic Equilibrium - 2 - Question 9

Which one is the correct expression below for the solution containing 'n' number of weak acids?

Detailed Solution for Test: Ionic Equilibrium - 2 - Question 9

Test: Ionic Equilibrium - 2 - Question 10

The pH of glycine at the first half equivalence point is 2.34 and that at second half equivalence paint is 9.60. At the equivalence point (The first inflection point) The pH is :

Detailed Solution for Test: Ionic Equilibrium - 2 - Question 10

Test: Ionic Equilibrium - 2 - Question 11

A 1.458 g of Mg reacts with 80.0 ml of a HCI solution whose pH is -0.477. The change in pH after all Mg has reacted. (Assume constant volume. Mg = 24.3 g/mol.)(log 3 = 0.477)

Detailed Solution for Test: Ionic Equilibrium - 2 - Question 11

Mg (aq) + 2HCl(aq) → MgCl2 (aq) + H

1.458/24.3

Millimoles of HCl = 3 x 80 = 240 mM
Moles of HCL after reaction = 240 - 60 x

New Molarity = 120/80 = 1.5 M

pH = -log[H+] = -log 1.5 = -0.176

change is pH = -0.176 - (-0.477) = 0.3

Test: Ionic Equilibrium - 2 - Question 12

The ionization constant of benzoic acid is 6.46 x 10-5 and lc for silver benzoate is 2.5 x 10-13. How many times silver benzoate more soluble in a buffer of pH = 3.19 as compared to its solubility in pure water?               

Detailed Solution for Test: Ionic Equilibrium - 2 - Question 12

The correct answer is Option A.
 
Ka of PhCOOAg= 6.45×10-5
Ksp of PhCOOAg= 2.5×10-13
pH of buffer = 3.19
∴[H] =Antilog(−3.19)
          =Antilog(−3−0.19+1−1)
          =Antilog(4.81)
          =6.45×10-4 M
Let Sbuffer(solubility of salt in a buffer)

=[1+10]1/2
=(11)1/2
=3.317
Thus, silver benzoate is 
3.317 times more soluble at lower 
pH than in water.
 

Test: Ionic Equilibrium - 2 - Question 13

A certain acid-base indicator is red in acid solution and blue in basic solution 75% of the indicator is present in the solution in its blue form at pH = 5. Calculate the pH at which the indicator shows 90% red form?                       

Detailed Solution for Test: Ionic Equilibrium - 2 - Question 13

Test: Ionic Equilibrium - 2 - Question 14

Which of the following concentrations of NH4+ will be sufficient to prevent the precipitation of Mg(OH)2 from a solution which is 0.01 M MgCl2 and 0.1 M NH3 (aq).

Given that : K5 of Mg (OH)2 = 2.5 x 10-11 and Kb for NH3(aq) = 2 x 10-6.                                               

Detailed Solution for Test: Ionic Equilibrium - 2 - Question 14

The expression for the solubility product of magnesium hydroxide is Ksp
[Mg2+][OH-]2
Thus 2.5×10-11 = 0.01 × [OH-]2
[OH-] = 5×10-5
The expression for dissociation constant of base is Kb = [NH4+][OH-]/[NH3]
Thus = 2×10-5 = [NH4+](5×10-5)/0.1
Hence, [NH4+] = 0.04 M

Test: Ionic Equilibrium - 2 - Question 15

An acid-base indicator which is a weak acid has a pKa value = 5.45. At what cocentration ratio of sodium acetate to acetic acid would the indicator show a colour half-way between those of its acid and conjugate base forms? pKa of acetic acid = 4.75. [log 2 = 0.3]                       

Detailed Solution for Test: Ionic Equilibrium - 2 - Question 15

Test: Ionic Equilibrium - 2 - Question 16

The indicator constant of phenolphthalein is approximately 10-10. A solution is prepared by adding 100.01 c.c. of 0.01 N sodium hydroxide to 100.00 c.c. of 0.01N hydrochloric acid. If a few drops of phenolphthalein are now added, what fraction of the indicator is converted to its coloured form?                       

Detailed Solution for Test: Ionic Equilibrium - 2 - Question 16

Test: Ionic Equilibrium - 2 - Question 17

A buffer solution is made by mixing a weak acid HA (Ka = 10-6) with its salt NaA in equal amounts. What should be the amount of acid or salt that should be added to make 90 ml of buffer solution of buffer capacity. 0.1 ? 

Detailed Solution for Test: Ionic Equilibrium - 2 - Question 17

For buffer capacity of 0.1 we should have

where is concentration of weak acid or salt in the buffer solution 

So, c+0.1 = 10c-1,

so 9c = 1.1 or c= 1.1/9

So, moles required for 90 solution = moles = 11 m moles

Test: Ionic Equilibrium - 2 - Question 18

 Statement-1 : solubility of BaSO4 in 0.1 M Na2SO4 is 10-9 M hence its K8 is 10-18.

Statement-2 : because for BaSO4 Ksp = (s)2 [symbols have their usual meanings].

Detailed Solution for Test: Ionic Equilibrium - 2 - Question 18

Test: Ionic Equilibrium - 2 - Question 19

                              

Statement-1 : It is difficult to distinguish the strengths of the strong acids such as HCI, H2SO4, HNO3,HBr, HI or HCIO4 in dilute aqueous solutions.

Statement-2 : In dilute aqueous solution all strong acids donate a proton to water and are essentially. 100% ionised to produce a solution containing H30+ ions plus the anions of strong acid .

Detailed Solution for Test: Ionic Equilibrium - 2 - Question 19

All are strong (α = 1) (leavelling effect).

Test: Ionic Equilibrium - 2 - Question 20

The degree of dissociation of water in a 0.1 M aqueous solution of HC1 at a cedain temperature t°C is 3.6 x 10-15. The temperature t must be :                                               

Detailed Solution for Test: Ionic Equilibrium - 2 - Question 20

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