   Chapter 15, Problem 121AP ### Introductory Chemistry: A Foundati...

9th Edition
Steven S. Zumdahl + 1 other
ISBN: 9781337399425

#### Solutions

Chapter
Section ### Introductory Chemistry: A Foundati...

9th Edition
Steven S. Zumdahl + 1 other
ISBN: 9781337399425
Textbook Problem
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# Calcium carbonate, CaCO3, can be obtained in a very pure state. Standard solutions of calcium ion are usually prepared by dissolving calcium carbonate in acid. What mass of CaCO3 should be taken to prepare 500. mL of 0.0200 M calcium ion solution?

Interpretation Introduction

Interpretation:

The mass of CaCO3 that should be taken to prepare the given calcium ion solution is to be calculated.

Concept Introduction:

The atomic mass of an element is defined as the sum of number of protons and number of neutrons. Molar mass of an element is determined from atomic mass of an element.

The number of moles is calculated by the formula,

Moles=MassgMolarmass

The molarity is calculated by the formula,

Molarity=NumberofmolesofsoluteVolumeofsolutionL.

Explanation

The molarity and volume of the calcium ion solution is given to be 0.0200M and 500.mL respectively.

The conversion of units of 500.mL into L is done as,

500.mL=500.1000L=0.5L

The equation of dissociation of CaCO3 is shown below.

CaCO3sH2OlCa2+aq+CO32

The above equation indicates that one mole of CaCO3 gives one mole of Ca2+ ions. Hence, the molarity of CaCO3 is equal to the molarity of Ca2+ ions.

The number of moles of CaCO3 is calculated by the formula,

Moles=Molarity×Volume        (1)

Substitute the values of molarity and volume of CaCO3 solution in the equation (1)

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