   Chapter 15, Problem 64QAP ### 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
21 views

# 64. Generally only the carbonates of the Group I elements and the ammonium ion are soluble in water: most other carbonates are insoluble. How many milliliters of 0.1 25 M sodium carbonate solution would be needed to precipitate the calcium ion from 37.2 mL of 0.105 M CaC12 solution? Na 2 CO3 ( aq ) + CaCl2 ( aq ) → CaCO3 ( s ) + 2NaCI ( aq )

Interpretation Introduction

Interpretation:

The milliliters of 0.125M sodium carbonate solution which would be required to precipitate all the calcium ions in a given solution are 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

Molarity is used to find out the concentration of solution.

The molarity is calculated by the formula,

Molarity=NumberofmolesofsoluteVolumeofsolutionL.

Explanation

The molarity and volume of CaCl2 solution is given to be 0.105M and 37.2mL respectively.

The conversion of units of 37.2mL into L is done as,

37.2mL=37.21000L=0.0372L

The number of moles of Ca2+ ions is calculated by the formula,

NumberofmolesofCa2+ions=Molarity×VolumeofsolutionL        (1)

Substitute the values of molarity and volume of CaCl2 solution in the equation (1).

NumberofmolesofCa2+ions=0.105M×0.0372L=0.003906moles

The equation for precipitation reaction is shown below.

Na2CO3aq+CaCl2aqCaCO3s+2NaClaq

It indicates that 0

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