   # A solution is prepared by mixing 100.0 mL of 1.0 × 10 −4 M Be(NO 3 ) 2 and 100.0 mL of 8.0 M NaF. Be 2+ ( a q ) + F − ( a q ) ⇌ B e F + ( a q ) K 1 = 7.9 × 10 4 BeF + ( a q ) + F − ( a q ) ⇌ B e F 2 ( a q ) K 2 = 5.8 × 10 3 BeF 2 ( a q ) + F − ( a q ) ⇌ B e F 3 − ( a q ) K 3 = 6.1 × 10 2 BeF 3 − ( a q ) + F − ( a q ) ⇌ B e F 4 2 − ( a q ) K 4 = 2.7 × 10 1 Calculate the equilibrium concentrations of F − , Be 2+ , BeF + , BeF 2 , BeF 3 − , and BeF 4 2− in this solution. ### Chemistry: An Atoms First Approach

2nd Edition
Steven S. Zumdahl + 1 other
Publisher: Cengage Learning
ISBN: 9781305079243

#### Solutions

Chapter
Section ### Chemistry: An Atoms First Approach

2nd Edition
Steven S. Zumdahl + 1 other
Publisher: Cengage Learning
ISBN: 9781305079243
Chapter 15, Problem 70E
Textbook Problem
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## A solution is prepared by mixing 100.0 mL of 1.0 × 10−4 M Be(NO3)2 and 100.0 mL of 8.0 M NaF. Be 2+ ( a q )   +   F − ( a q )   ⇌   B e F + ( a q )   K 1   =   7.9   ×   10 4 BeF + ( a q )   +   F − ( a q )   ⇌   B e F 2 ( a q )   K 2   =   5.8   ×   10 3 BeF 2 ( a q )   +   F − ( a q )   ⇌   B e F 3 − ( a q ) K 3   =   6.1   ×   10 2 BeF 3 − ( a q )   +   F − ( a q )   ⇌   B e F 4 2 − ( a q )   K 4   =   2.7   ×   10 1 Calculate the equilibrium concentrations of F−, Be2+, BeF+, BeF2, BeF3−, and BeF42− in this solution.

Interpretation Introduction

Interpretation: The value of different equilibrium constant for the reaction involving the formation of BeF42 in a stepwise manner is given. The equilibrium concentration of BeF3 , BeF42 , BeF42 , BeF2 , BeF+ , Be2+ is to be calculated.

Concept introduction: The equilibrium that is present between the species that is unionized and its ions is called ionic equilibrium.

### Explanation of Solution

Explanation

To determine: The equilibrium concentration of BeF3 , BeF42 , BeF42 , BeF2 , BeF+ , Be2+ in the given solution.

Explanation

The equilibrium concentration of BeF42 is 5.0×105 and equilibrium concentration of F is 4.0M .

Given

The concentration of Be(NO3)2 is 1.0×104M .

The concentration of NaF is 8.0M .

The volume of Be(NO3)2 is 100.0mL .

The volume of NaF is 100.0mL .

The value of K1 is 7.9×104 .

The value of K2 is 5.8×103 .

The value of K3 is 6.1×102 .

The value of K4 is 2.7×101 .

The concentration of any species in the solution before any reaction takes place is given as,

MiVi=MfVf

Where,

• Mi is the initial molarity.
• Mf is the final molarity.
• Vi is the initial volume of the solution.
• Vf is the volume of solution after mixing.

Substitute the value of Mi , Vi and Vf of Be2+ in the above equation as,

MiVi=MfVf1.0×104M×100mL=Mf×200mLMf=5.0×105M

Similarly substitute the value of Mi , Vi and Vf of F in the above equation as,

MiVi=MfVf8.0M×100mL=Mf×200mLMf=4.00M

As it is seen from the given values of equilibrium constants that the reaction almost goes towards the completion. The final equation is therefore given as,

Be2++4FBeF42Before reaction(M) 5.0×1054.000Change (M)  5.0×1055.0×105+5.0×105After reaction (M)04.004(5.0×105)5.0×105

This gives the equilibrium concentration of BeF42 as 5.0×10-5_ and equilibrium concentration of F as 4.0M_ .

The equilibrium concentration of [BeF3] is 4.6×10-7M_ .

The equilibrium constant for the reaction involving the formation of BeF42 is given as,

K4=[BeF42][BeF3][F]

Substitute the values of K4 , [BeF42] and [F] in the above equation as,

K4=[BeF42][BeF3][F]2.7×101=5.0×105M[BeF3](4.004(5.0×105))M[BeF3]=5.0×105M2.7×101×(4.004(5.0×105))M=4

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