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JEE Exam  >  JEE Tests  >  Daily Test for JEE Preparation  >  Test: Ionic Equilibrium in Solution (24 July) - JEE MCQ

Test: Ionic Equilibrium in Solution (24 July) - JEE MCQ


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10 Questions MCQ Test Daily Test for JEE Preparation - Test: Ionic Equilibrium in Solution (24 July)

Test: Ionic Equilibrium in Solution (24 July) for JEE 2024 is part of Daily Test for JEE Preparation preparation. The Test: Ionic Equilibrium in Solution (24 July) questions and answers have been prepared according to the JEE exam syllabus.The Test: Ionic Equilibrium in Solution (24 July) MCQs are made for JEE 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: Ionic Equilibrium in Solution (24 July) below.
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Test: Ionic Equilibrium in Solution (24 July) - Question 1
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Out of the following, amphiprotic species are
I. HPO32-
II. OH-
III. H2PO4-
IV. HCO3-

Detailed Solution for Test: Ionic Equilibrium in Solution (24 July) - Question 1

Amphiprotic : can accept and Release H+
Only H2PO–4 & HCO–3

Test: Ionic Equilibrium in Solution (24 July) - Question 2
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1 c.c. of 0.1N HCl is added to 99 CC solution of NaCl. The pH of the resulting solution will be

Detailed Solution for Test: Ionic Equilibrium in Solution (24 July) - Question 2

Volume of resulting solution = 100 ml

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Test: Ionic Equilibrium in Solution (24 July) - Question 3
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The pH of an aqueous solution of 1.0 M solution of a weak monoprotic acid which is 1% ionised is:

Detailed Solution for Test: Ionic Equilibrium in Solution (24 July) - Question 3

Concentration of solution = C = 1M
Dissociation is α = 0.01
After dissociation, the concentration of H+ ions will be Cα = 1 × 0.01 = 10−2
pH = −log[H+] = −log(10−2) = +2

Test: Ionic Equilibrium in Solution (24 July) - Question 4
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What is the percentage hydrolysis of NaCN in N/80 solution when the dissociation constant for HCN is 1.3 × 10-9 and Kw = 1.0 × 10-14
Detailed Solution for Test: Ionic Equilibrium in Solution (24 July) - Question 4

Weak acid + Strong base : Salt


Test: Ionic Equilibrium in Solution (24 July) - Question 5
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The ≈ pH of the neutralisation point of 0.1 N ammonium hydroxide with 0.1 N HCl is
Detailed Solution for Test: Ionic Equilibrium in Solution (24 July) - Question 5

Salt formed : NH4Cl = 0.1 N
Solution will be slightly acidic due to Hydrolysis

Test: Ionic Equilibrium in Solution (24 July) - Question 6
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A solution with pH 2.0 is more acidic than the one with pH 6.0 by a factor of :
Detailed Solution for Test: Ionic Equilibrium in Solution (24 July) - Question 6

[H+]1 = 10–2 ; [H+]2 = 10–6

Test: Ionic Equilibrium in Solution (24 July) - Question 7
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What volume of 0.2 M NH4Cl solution should be added to 100 ml of 0.1 M NH4OH solution to produce a buffer solution of pH = 8.7 ?            

Given : pKb of NH4OH = 4.7 ; log 2 = 0.3
Detailed Solution for Test: Ionic Equilibrium in Solution (24 July) - Question 7

pH = 8.7 ⇒ pOH = 5.3
Basic Buffer

If volume of salt = V ml

Test: Ionic Equilibrium in Solution (24 July) - Question 8
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The pKa of a weak acid, HA, is 4.80. The pKb of a weak base, BOH, is 4.78. The pH of an aqueous solution of the corresponding salt, BA, will be:
Detailed Solution for Test: Ionic Equilibrium in Solution (24 July) - Question 8

Salt of weak acid & weak base

Test: Ionic Equilibrium in Solution (24 July) - Question 9
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Which of the following in most soluble in water ?
Detailed Solution for Test: Ionic Equilibrium in Solution (24 July) - Question 9

Calculate the solubility ‘s’ for each option, Higher the value of ‘s’ Higher the solubility.

Test: Ionic Equilibrium in Solution (24 July) - Question 10
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pH of saturated solution of silver salt of monobasic acid HA is found to be 9.

Find the Ksp of sparingly soluble salt Ag A(s).

Given : Ka(HA) = 10-10
Detailed Solution for Test: Ionic Equilibrium in Solution (24 July) - Question 10




Ksp  = S (S–X) = 11 × 10–6 × 10–6
= 1.1 × 10–11

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