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A solution containing 4.2 g of KOH and Ca(OH)2 is neutralised by and acid.It consumes .1 equivalent of acid ,calculate equimolar amounts of the sample.?
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A solution containing 4.2 g of KOH and Ca(OH)2 is neutralised by and a...
First of all, always remember that 1 equivalent of an acid always reacts with 1 equivalent of a base. It doesn’t matter which acid or base we’re taking.
Step 1: First let’s find the equivalent weights of KOH and Ca(OH)2
(a) KOH: Molar mass of KOH= 39 + 16 + 1 = 56. Valency factor of KOH = 1. So, equivalent weight of KOH = 56
(b) Ca(OH)2: Molar mass of Ca(OH)2= 40 + 2*(16+1) = 74. Valency factor of Ca(OH)2= 2. So, equivalent weight = 74/2 = 37
Step 2: We’ve been given 0.1 equivalent of acid. So, some of this will react with KOH and some will react with Ca(OH)2
Let the number of equivalents of acid that react with Ca(OH)2 = x. So, the number of equivalents of acid that react with KOH = 0.1 - x
Step 3: Let the mass of Ca(OH)2 = y, and so the mass of KOH= 4.2 -y
Step 4: Time for some math. 1 equivalent of acid will react with 1 equivalent of Ca(OH)2. That means 1 equivalent of acid will react with 37 g of Ca(OH)2.
So, by proportion, 1/37 = x/y
or, y = 37x
Similarly for KOH, 1/56= 0.1-x/4.2-y
Putting value of y as 37 x and cross multiplying
4.2–37x=5.6–56x
x= 0.0737
So, y = 37x = 37*0.0737 = 2.73 g (approx)
so, 4.2 - y = 4.2 - 2.73 = 1.47 g (approx)
So, mass of Ca(OH)2 = 2.73 g and mass of KOH= 1.47g
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Most Upvoted Answer
A solution containing 4.2 g of KOH and Ca(OH)2 is neutralised by and a...
Given:
- Amount of KOH = 4.2 g
- Acid consumed = 0.1 equivalents

To Find:
- Equimolar amounts of the sample

Solution:

Step 1: Calculate the Molar Mass
- Molar mass of KOH = 39.1 g/mol (K) + 16.0 g/mol (O) + 1.0 g/mol (H) = 56.1 g/mol
- Molar mass of Ca(OH)2 = 40.1 g/mol (Ca) + 2 * (16.0 g/mol (O) + 1.0 g/mol (H)) = 74.1 g/mol

Step 2: Calculate the Moles
- Moles of KOH = Mass of KOH / Molar mass of KOH = 4.2 g / 56.1 g/mol ≈ 0.075 mol
- Moles of Ca(OH)2 = 0.075 mol (Since it is equimolar to KOH)

Step 3: Calculate the Moles of Acid
- Moles of acid consumed = 0.1 equivalents

Step 4: Calculate the Equimolar Amounts
- Since the moles of KOH and Ca(OH)2 are equal, the equimolar amounts of the sample are:
- KOH: 0.075 mol
- Ca(OH)2: 0.075 mol

Explanation:
- The molar mass of KOH is calculated by adding the atomic masses of potassium (K), oxygen (O), and hydrogen (H). Similarly, the molar mass of Ca(OH)2 is calculated by adding the atomic masses of calcium (Ca), oxygen (O), and hydrogen (H).
- The moles of KOH and Ca(OH)2 are determined by dividing the given mass of each compound by their respective molar masses.
- Since the acid consumed is given in equivalents, it represents the number of moles of acid required to neutralize the solution.
- Finally, since KOH and Ca(OH)2 are equimolar in the given solution, the equimolar amounts of the sample are equal to the moles of KOH and Ca(OH)2, which is 0.075 mol for each compound.
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A solution containing 4.2 g of KOH and Ca(OH)2 is neutralised by and acid.It consumes .1 equivalent of acid ,calculate equimolar amounts of the sample.?
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A solution containing 4.2 g of KOH and Ca(OH)2 is neutralised by and acid.It consumes .1 equivalent of acid ,calculate equimolar amounts of the sample.? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about A solution containing 4.2 g of KOH and Ca(OH)2 is neutralised by and acid.It consumes .1 equivalent of acid ,calculate equimolar amounts of the sample.? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A solution containing 4.2 g of KOH and Ca(OH)2 is neutralised by and acid.It consumes .1 equivalent of acid ,calculate equimolar amounts of the sample.?.
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