JEE Main 2019ChemistryIonic EquilibriumPh Buffer And IndicatorsmediumMCQ

JEE Main 2019Ionic Equilibrium Question with Solution

From: JEE Main 2019 (Online) 10th April Morning Slot

Question

Consider the following statements
(a) The pH of a mixture containing 400 mL of 0.1 M H2SO4 and 400 mL of 0.1 M NaOH will be approximately 1.3
(b) Ionic product of water is temperature dependent.
(c) A monobasic acid with Ka = 10–5 has pH = 5. The degree of dissociation of this acid is 50 %.
(d) The Le Chatelier's principle is not applicable to common-ion effect.

The correct statements are :

Choose an option

Show full solutionCorrect option: B
Correct answer
B(a), (b) and (c)

Step-by-step explanation

(a) Equivalance of strong acid = 0.1 2 400 = 80

Equivalance of strong base = 0.1 400 = 40

[H+] of mixture = =

pH = = = 1.3

(b) Ionic product of water increases with increase of temperature because ionisation of water is endothermic.

(c) JEE Main 2019 (Online) 10th April Morning Slot Chemistry - Ionic Equilibrium Question 90 English Explanation

ka =

10-5 =

= 1

= 0.5

Degree of dissociation() = 50%

(d) The Le Chatelier's principle is always applicable to common-ion effect.

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About this question

This is a previous-year question from JEE Main 2019, covering the Ionic Equilibrium chapter of Chemistry. PrepSharp catalogues every PYQ from JEE Main with a verified answer key and step-by-step solution prepared by IIT alumni — so you can search by chapter, topic or year and revise efficiently.