Explanation Given information: The mass of the mixture of NaOH and Ba(OH) 2 is 10.0 g. It is dissolved in 250.0 mL of water. The molarity of HCl is 1.50 M. The volume of acid added is 108.9 mL. Calculation: A kind of technique used to determine the concentration of an unknown solution with the help of a solution whose concentration is known. This technique is also known as volumetric analysis as it as the best care of volume is being taken while doing the titration. The formula used in the titration to determine the concentration of other solution is: M 1 V 1 = M 2 V 2 Where, M 1 , V 1 is the molarity and volume of the first solution which is taken as standard and M 2 , V 2 is the molarity and volume of the second solution whose concentration is to be measured. For determining the mass of each component in the mixture, start with the number of moles of OH - . H + + OH − → H 2 O Moles of H+ = molarity x litre Moles of H+ = 1.50 x 0.1089 Hence, moles of H+ = 0.1633 ..... (1) Total number of OH- = total number of H+ Total number of OH- = 0.1633 moles Let the number of moles of NaOH in the solution be X The number of moles of Ba(OH) 2 in the solution be Y It is clear from equation (1) that 1 mole of OH- gives 1 mole of NaOH and 2 moles of Ba(OH) 2 Hence, X + 2 Y = 0.1633 X = 0.1633 – 2 Y ..... (2) Also, it is given in question that total mass of the mixture is equal to 10gm Therefore, M NaOH + M Ba ( OH ) 2 = 10 g

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ISBN: 9780135216972

Author: Robinson

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Acid Base Titration

An acid-base titration is the method of quantitative analysis for determining the concentration of an acid and base by exactly neutralizing it with a standard solution of base or acid having known concentration.

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Chapter 4, Problem 4.152MP

Interpretation Introduction

Interpretation:

The mass of each substance in the mixture needs to be determined.

Concept introduction:

The number of moles is related to mass and molar mass as follows:

n=mM

Here, m is mass and M is molar mass.

The molarity of the solution is defined as the number of moles of solute dissolved in 1 L of the volume of solution.

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Chapter 4, Problem 4.151MP

Chapter 4, Problem 4.153MP

Chapter 4 Solutions

EBK CHEMISTRY

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Solution Summary: The author explains that the mass of each substance in the mixture needs to be determined.

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