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Determine the volume of 0.225 M KOH solution required to

Introductory Chemistry | 5th Edition | ISBN: 9780321910295 | Authors: Nivaldo J Tro ISBN: 9780321910295 34

Solution for problem 90P Chapter 13

Introductory Chemistry | 5th Edition

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Introductory Chemistry | 5th Edition | ISBN: 9780321910295 | Authors: Nivaldo J Tro

Introductory Chemistry | 5th Edition

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Problem 90P

Determine the volume of 0.225 M KOH solution required to neutralize each sample of sulfuric acid. The neutralization reaction is:

H2SO4(aq) + 2 KOH(aq) → K2SO4(aq) + 2H2O(l)

(a) 45 mL of 0.225 M H2SO4

(b) 185 mL of 0.125 M H2SO4

(c) 75 mL of 0.100 M H2SO4

Step-by-Step Solution:
Step 1 of 3

Solution: Here, we are going to calculate the volume of KOH solution required to neutralize the given H2SO4 solutions.

Step1:

Neutralisation reaction is the reaction between an acid and a base to give salt and water. The neutralization reaction in the given problem is:

                                2KOH(aq) + H2SO4(aq) → 2H2O(l) + K2SO4(aq)

Here, 1 mole of H2SO4 requires 2 moles of KOH for complete neutralization.

Step2:

Again, molarity(M) is defined as number of moles of solute dissolved in one litre (or one cubic decimetre) of solution.

                                                   Moles of solute

                                Molarity = -------------------------------          ------(1)

                                              Volume of solution in litre

Step3:

Given, volume of H2SO4 solution = 45 mL = 45/1000 L = 0.045 L        [1000 mL = 1 L]

                Molarity of H2SO4 solution = 0.225 M

        Therefore, number of moles of H2SO4 = Molarity x volume of solution in litres

                                                = 0.225 M x 0.045 L

                                                = 0.010125 mol

Step4:

Number of moles of KOH = 2 x number of moles of H2SO4 = 2 x 0.010125 mol = 0.02025 mol

Molarity of KOH solution = 0.225 M

Therefore, volume of KOH solution required = moles of KOH / molarity

                                                  = 0.02025 mol / 0.225 M

                                                   = 0.09 L

                                                   = 0.09 x 1000 mL                [1 L = 1000 mL]

                                                   = 90 mL

Thus, 90 mL of KOH solution will be required.

Step5:

   b)        Given, volume of H2SO4 solution = 185 mL = 185/1000 L = 0.185 L     [1000 mL = 1 L]

                Molarity...

Step 2 of 3

Chapter 13, Problem 90P is Solved
Step 3 of 3

Textbook: Introductory Chemistry
Edition: 5
Author: Nivaldo J Tro
ISBN: 9780321910295

The full step-by-step solution to problem: 90P from chapter: 13 was answered by , our top Chemistry solution expert on 05/06/17, 06:45PM. Introductory Chemistry was written by and is associated to the ISBN: 9780321910295. This full solution covers the following key subjects: koh, reaction, acid, neutralization, neutralize. This expansive textbook survival guide covers 19 chapters, and 2045 solutions. Since the solution to 90P from 13 chapter was answered, more than 1506 students have viewed the full step-by-step answer. This textbook survival guide was created for the textbook: Introductory Chemistry, edition: 5. The answer to “Determine the volume of 0.225 M KOH solution required to neutralize each sample of sulfuric acid. The neutralization reaction is:H2SO4(aq) + 2 KOH(aq) ? K2SO4(aq) + 2H2O(l)(a) 45 mL of 0.225 M H2SO4(b) 185 mL of 0.125 M H2SO4(c) 75 mL of 0.100 M H2SO4” is broken down into a number of easy to follow steps, and 45 words.

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