INTRO. TO CHEM LOOSELEAF W/ALEKS 18WKCR
INTRO. TO CHEM LOOSELEAF W/ALEKS 18WKCR
5th Edition
ISBN: 9781264125609
Author: BAUER
Publisher: MCG
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Chapter 11, Problem 80QP

(a)

Interpretation Introduction

Interpretation:

The volume of 0.5000 MHCl required to neutralize 35.00mL of 0.0500 MNaOH is to be calculated.

(a)

Expert Solution
Check Mark

Explanation of Solution

The balanced chemical equation is,

HCl(aq)+NaOH(aq)NaCl(aq)+H2O(l)

The moles of NaOH present in 35.00mL of 0.0500 MNaOH after converting millilitre to litre unit is calculated as follows,

Moles of  NaOH=35.00mL×1L1000mL×0.0500mol/L=0.00175mol

From the balanced chemical reaction, it is clear that 1mol of HCl reaction with 1mol of NaOH . The mol of HCl is calculated using the mol ratio as shown below.

Moles of HCl=0.00175molNaOH×1mol HCl1mol NaOH=0.00175mol HCl

The relationship between molarity, moles of solute, and volume of the solution is,

Molarity=Moles of soluteVolume of solution in litre    ......1

Equation 1 can be rearranged to the following equation.

Volume in litre=Moles of soluteMolarity

The volume of HCl required is,

Volume of HCl in litre=0.00175mol0.5000mol L1=0.0035L

The required volume in liter unit is very small. Thus, it must be converted to milliliter unit.

Volumeof NaOH in millilitre=0.0035L×1000mL1L=3.50mL

Hence, 3.50mL of 0.0500 MNaOH is required to neutralize 35.0mL of 0.5000 MHCl .

(b)

Interpretation Introduction

Interpretation:

The volume of 0.5000 MHCl required to neutralize 10.00mL of 0.2000 MBa(OH)2 is to be calculated.

(b)

Expert Solution
Check Mark

Explanation of Solution

The balanced chemical equation is,

Ba(OH)2(aq)+2HCl(aq)BaCl2(aq)+2H2O(l)

The moles of Ba(OH)2 present in 15.00mL of 0.2000 MBa(OH)2 after converting milliliter into liter unit is calculated as follows,

Moles of Ba(OH)2=15.00mL×1L1000mL×0.2000mol L1=0.003mol

From the balanced chemical reaction, it is clear that 2moles of HCl react with 1mol of Ba(OH)2 . The moles of HCl is calculated using the mol ratio as shown below.

Moles of HCl=0.003mol Ba(OH)2×2mol HCl1mol Ba(OH)2=0.006mol HCl

The relationship between molarity, moles of solute, and volume of the solution is,

Molarity=Moles of soluteVolume of solution in litre    ......1

Equation 1 can be rearranged to the following equation.

Volume in litre=Moles of soluteMolarity

The volume of HCl required is as shown below.

Volumeof HCl in litre=0.006mol0.5000mol L1=0.012L

The required volume in liter unit is very small. Thus, it must be converted to milliliter.

Volumeof HCl in millilitre=0.012L×1000mL1L=12.0mL

Hence, 12.0mL of 0.5000 MHCl required to neutralize 15.00mL of 0.2000 MBa(OH)2 .

(c)

Interpretation Introduction

Interpretation:

The volume of 0.5000 MHCl required to neutralize 15.00mL of 0.2500MNH3 is to be calculated.

(c)

Expert Solution
Check Mark

Explanation of Solution

The balanced chemical equation is,

HCl(aq)+NH3(aq)NH4Cl(aq)

The moles of NH3 present in 15.00mL of 0.2500 MNH3 after converting milliliter to liter unit is calculated as follows,

Moles NH3=15.00mL×1L1000mL×0.2500mol L1=0.00375mol

From the balanced chemical reaction, it is clear that 1mol of HCl reacts with 1mol of NH3 . The moles of HCl is calculated using the mol ratio as shown below.

Moles HCl=0.00375mol NH3×1mol HCl1mol NH3=0.00375mol HCl

The relationship between molarity, moles of solute, and volume of the solution is,

Molarity=Moles of soluteVolume of solution in litre    ......1

Equation 1 can be rearranged to the following equation.

Volume in litre=Moles of soluteMolarity

The volume of HCl required is as given below.

Volumeof HCl in litre=0.00375mol0.5000mol L1=0.0075L

The required volume in liter unit is very small. Thus, it must be converted into milliliters.

Volumeof HCl in mollilitre=0.0075L×1000mL1L=7.5mL

Hence, 7.5mL of 0.5000 MHCl required to neutralize 15.00mL of 0.2500MNH3 .

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Chapter 11 Solutions

INTRO. TO CHEM LOOSELEAF W/ALEKS 18WKCR

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