General Chemistry
General Chemistry
7th Edition
ISBN: 9780073402758
Author: Chang, Raymond/ Goldsby
Publisher: McGraw-Hill College
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Chapter 16, Problem 16.98QP
Interpretation Introduction

Interpretation:

The concentration of all species in a 0.100 M Na2CO3 solution has to be calculated

Concept Information:

Base  ionization constant Kb

The equilibrium expression for the ionization of weak base B will be,

  B(aq)+H2O(l)HB+(aq)+OH-(aq)

  Kb=[HB+][OH-][B]

Where Kb is base ionization constant, [OH] is concentration of hydroxide ion, [HB+] is concentration of conjugate acid, [B] is concentration of the base

To Calculate: The concentration of all species in a 0.100 M Na2CO3 solution

Expert Solution & Answer
Check Mark

Answer to Problem 16.98QP

The concentration of all species in a 0.100 M Na2CO3 solution is:

  [Na+] =0.200 M[CO32-] =0.095M[HCO3-] =4.6×103M[H2CO3] =2.4×10-8M[OH-] =4.6×103M[H+] =2.2×1012M

Explanation of Solution

Given data:

The concentration of Na2CO3 solution is 0.100 M

Ionization of Na2CO3

Na2CO3 is a weak base with a concentration of 0.100 M.

0.100MNa2CO30.200MNa++0.100MCO32-

Using the ionization constant for Na2CO3 the concentration of Na+, HCO3-,CO32,H2CO3, OH- and H+ can be calculated as follows.

First stage:

The concentration of all the species before and after ionization can be represented as follows,

 CO32(aq)+H2O(l)HCO3(aq)+OH-(aq)
Initial (M)

0.100

x

0.100-x

0.000.00
Change (M)+x+x
Equilibrium (M)xx

The Ka for HCO3 is 4.8×10-11

  K1 =KwK2 =1.0×10144.8×1011 =2.1×104

       K1 =[HCO3-][OH-][CO32-]   2.1×104 =x2(0.100x) x20.100 x =4.6×103M[HCO3-]=[OH-]=4.6×103M

Therefore, the concentration of [HCO3-]=[OH-]=4.6×103M

Second stage:

Let y be the change in concentration.

Note the initial concentration of [HCO3-] is 4.6×103M from the first ionization.

 HCO3(aq)+H2O(l)H2CO3(aq)+  OH(aq)
Initial (M)

4.6×103

y

(4.6×103)y

0.000.00
Change (M)+y+y
Equilibrium (M)y(4.6×103)+y

   K2=[H2CO3][OH-][HCO3-]2.4×108=y[(4.6×103)+y](4.6×103)y(4.6×103)(y)4.6×103 y=2.4×108M

At equilibrium:

  [Na+] =0.200 M[CO32-] =(0.100-4.6×10-3)M =0.095M[HCO3-]=(4.6×10-3)M(2.4×108)M 4.6×103M[H2CO3]=2.4×10-8M[OH-] =(4.6×10-3)M+(2.4×108)M 4.6×103M[H+] =1.0×10144.6×103 =2.2×1012M

Conclusion

The concentration of all species in a 0.100 M Na2CO3 solution was calculated

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Chapter 16 Solutions

General Chemistry

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