JEE Main 2021
LEVELJEE Advanced
Visualized Solution
The Sigma Insight: Stoichiometric and Volumetric Calculations
Welcome to one of the most elegant techniques in analytical chemistry: Double Titration. This method is a brilliant piece of chemical detective work that allows us to find the exact composition of a mixture containing multiple bases, simply by using two different indicators in sequence.
The Mystery Mixture
Imagine you are handed a conical flask containing exactly of a solid mixture. Inside this mixture, there is strong sodium hydroxide (), moderately basic sodium carbonate (), and some useless inert impurities. Your mission is to find out exactly what percentage of this powder is sodium carbonate.
To do this, we will titrate the mixture with a standard solution of hydrochloric acid (). The concentration of our acid is given as , which is equivalent to since the n-factor of is .
The First Milestone
Phenolphthalein
We begin by adding a few drops of phenolphthalein to our flask. The solution immediately turns a vibrant pink because the mixture is highly basic. As we drip from the burette, the acid starts neutralizing the bases.
Phenolphthalein is a very specific indicator; it changes color around a pH of . By the time the pH drops to this level, two things have happened:
1. The strong base, , has been completely neutralized into and water.
2. The has only been half-neutralized. It has grabbed one proton to become sodium bicarbonate ().
At this exact moment, the pink color vanishes. The burette tells us we have used of acid. This volume represents the complete neutralization of plus the half-neutralization of .
The Second Milestone
Methyl Orange
Now, we add methyl orange to the very same flask. The solution turns yellow. We continue adding .
What is the acid reacting with now? The is already gone. The only basic substance left is the that we just created in the first step. The acid attacks this bicarbonate, converting it fully into carbonic acid (), which breaks down into water and carbon dioxide.
When this reaction is complete, the pH drops further, and the methyl orange turns red. The problem states that this second step required exactly of acid.
Cracking the Code
The Math
This is our golden key. It represents the exact amount of acid needed to convert the intermediate into carbonic acid.
Let's calculate the moles of acid used in this second step:
Because the stoichiometry between and is , we know there were exactly of in the flask.
But where did this bicarbonate come from? It came entirely from the original sodium carbonate! Since one molecule of produces exactly one molecule of , the moles of our original sodium carbonate must also be .
Now, we convert these moles into mass. The molar mass of is .
This is equal to .
The Final Percentage
We have found the mass of the sodium carbonate. The final step is to find its weight percentage in the original mixture.
The question asks us to round off to the nearest integer. Rounding gives us our final, elegant answer: .
As a thought experiment, if you wanted to find the mass of , you would realize that the first included dedicated to the first half of the carbonate. Thus, the volume used exclusively for would be . Double titration is truly a masterpiece of logical deduction!
Similar Questions
JEE Main 2021
LEVELJEE Main
10.0 mL of solution is titrated against 0.2 M HCl solution. The following titre values were obtained in 5 readings. 4.8 mL, 4.9 mL, 5.0 mL, 5.0 mL and 5.0 mL based on these readings and convention of titrimetric estimation of concentration of solution is ……… mM (Round off to the nearest integer).
JEE Main 2020
LEVELJEE Main
A solution was made by adding of . The normality of the solution is . The value of is ......... .
JEE Main 2019
LEVELJEE Main
25 mL of the given HCl solution requires 30 mL of 0.1 M sodium carbonate solution. What is the volume of this HCl solution required to titrate 30 mL of 0.2 M aqueous NaOH solution?
(A)
75 mL
(B)
25 mL
(C)
12.5 mL
(D)
50 mL
JEE Main 2019
LEVELJEE Advanced
A effervescent tablet containing sodium bicarbonate and oxalic acid releases of at and . If molar volume of is under such condition, what is the percentage of sodium bicarbonate in each tablet? [Molar mass of ]
(A)
8.4
(B)
0.84
(C)
16.8
(D)
33.6
JEE Advanced 2020
LEVELJEE Main
5.00 mL of 0.10 M oxalic acid solution taken in a conical flask is titrated against NaOH from a burette using phenolphthalein indicator. The volume of NaOH required for the appearance of permanent faint pink color is tabulated below for five experiments. What is the concentration, in molarity, of the NaOH solution ?
JEE Main 2021
LEVELJEE Main
Consider titration of NaOH solution versus oxalic acid solution. At the end point following burette readings were obtained. (i) (ii) (iii) (iv) (v) If the volume of oxalic acid taken was , then the molarity of the NaOH solution is ......... M. (Rounded off to the nearest integer)
JEE Advanced 2021
LEVELJEE Advanced
Comprehension Passage
A sample (5.6 g) containing iron is completely dissolved in cold dilute HCl to prepare a 250 mL of solution. Titration of 25.0 mL of this solution requires 12.5 mL of 0.03 M KMnO4 solution to reach the end point. Number of moles of Fe2+ present in 250 mL solution is x × 10–2 (consider complete dissolution of FeCl2). The amount of iron present in the sample of y% by weight.
(Assume : KMnO4 reacts only with Fe2+ in the solution
Use : Molar mass of iron as 56 g mol–1)
Question 1:
The value of x is ______.
Question 2:
The value of y is ______.
JEE Main 2017
LEVELJEE Main
1 g of a carbonate () on treatment with excess HCl produces 0.01186 mole of . The molar mass of in is
(A)
1186
(B)
84.3
(C)
118.6
(D)
11.86
JEE Main 2020
LEVELJEE Main
The ammonia () released on quantitative reaction of urea () with sodium hydroxide () can be neutralised by
(A)
of
(B)
of
(C)
of
(D)
of
JEE Main 2019
LEVELJEE Main
50 mL of 0.5 M oxalic acid is needed to neutralise 25 mL of sodium hydroxide solution. The amount of NaOH in 50 mL of the given sodium hydroxide solution is
(A)
40 g
(B)
80 g
(C)
20 g
(D)
10 g