Explanation The given equation is written as follows, HIO 3 (aq) + NH 2 NH 2 (aq) ⇌ NH 2 NH 3 + (aq) + IO 3 − (aq) Now, write the expression for equilibrium constant from the above reaction. K = [ NH 2 NH 3 + ] [ IO 3 − ] [ HIO 3 ] [ NH 2 NH 2 ] The NH 2 NH 3 + (aq) ion formation reaction equation is the combination of the following two equations and it can be written as, HIO 3 (aq) + H 2 O (l) ⇌ H 3 O + (aq) + IO 3 − (aq) NH 2 NH 2 (aq) + H 2 O (l) ⇌ NH 2 NH 3 + (aq) + OH - (aq) The equations represents the weak acid dissociation and weak base dissociation. Now, write the expressions for both acid and base dissociation constants for the above two equation as below. K a = [ H 3 O + ] [ IO 3 - ] [ HIO 3 ] K b = [ NH 2 NH 3 + ] [ OH - ] [ NH 2 NH 2 ] At 298K K a of HIO 3 is 1 .7×10 -1 K b of NH 2 NH 2 is 1 .7×10 -6 . Now, substitute these values in the above expression and it becomes, For HIO 3 K a = [ H 3 O + ] [ IO 3 − ] [ HIO 3 ] For NH 2 NH 2 , K b = [ NH 2 NH 3 + ] [ OH - ] [ NH 2 NH 2 ] 1 .7×10 -1 = [ H 3 O + ] [ IO 3 − ] [ HIO 3 ] 1 .7×10 -6 = [ NH 2 NH 3 + ] [ OH - ] [ NH 2 NH 2 ] Now, multiply the above two equations

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Chapter 6, Problem 6.12E

Interpretation Introduction

Interpretation:

The equilibrium constant for the following reaction and the value of K at 298 K for the reaction HIO3(aq) + NH2NH2(aq) ⇌ NH2NH3+ (aq) + IO3− (aq) has to be calculated.

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Chapter 6, Problem 6.11E

Chapter 6, Problem 6.13E

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Chapter 6 Solutions

CHEM PRINCIPLES LL W/ACHIEVE ONE-SEM

Ch. 6 - Prob. 6A.1AST Ch. 6 - Prob. 6A.1BST Ch. 6 - Prob. 6A.2AST Ch. 6 - Prob. 6A.2BST Ch. 6 - Prob. 6A.3AST Ch. 6 - Prob. 6A.3BST Ch. 6 - Prob. 6A.1E Ch. 6 - Prob. 6A.2E Ch. 6 - Prob. 6A.3E Ch. 6 - Prob. 6A.4E

Knowledge Booster

The equilibrium constant for the following reaction and the value of K at 298 K for the reaction HIO 3 (aq) + NH 2 NH 2 (aq) ⇌ NH 2 NH 3 + (aq) + IO 3 − (aq) has to be calculated.

Solution Summary: The author explains the equilibrium constant and the value of K at 298 K for the reaction. The ion formation reaction equation is the combination of the following two equations.

Interpretation:

The equilibrium constant for the following reaction and the value of K at 298 K for the reaction HIO3(aq) + NH2NH2(aq) ⇌ NH2NH3+ (aq) + IO3− (aq) has to be calculated.

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Chapter 6 Solutions

The equilibrium constant for the following reaction and the value of K at 298 K for the reaction HIO 3 (aq) + NH 2 NH 2 (aq) ⇌ NH 2 NH 3 + (aq) + IO 3 − (aq) has to be calculated.

Solution Summary: The author explains the equilibrium constant and the value of K at 298 K for the reaction. The ion formation reaction equation is the combination of the following two equations.

Interpretation: The equilibrium constant for the following reaction and the value of K at 298 K for the reaction HIO3(aq) + NH2NH2(aq) ⇌ NH2NH3+ (aq) + IO3− (aq) has to be calculated.

Want to see the full answer?

Chapter 6 Solutions

Interpretation:

The equilibrium constant for the following reaction and the value of K at 298 K for the reaction HIO3(aq) + NH2NH2(aq) ⇌ NH2NH3+ (aq) + IO3− (aq) has to be calculated.