EBK GENERAL CHEMISTRY: THE ESSENTIAL CO
EBK GENERAL CHEMISTRY: THE ESSENTIAL CO
7th Edition
ISBN: 8220106637203
Author: Chang
Publisher: YUZU
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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

EBK GENERAL CHEMISTRY: THE ESSENTIAL CO

Ch. 16.4 - Prob. 1RCCh. 16.5 - Prob. 1PECh. 16.5 - Prob. 2PECh. 16.5 - Prob. 1RCCh. 16.5 - Prob. 3PECh. 16.5 - Prob. 2RCCh. 16.6 - Prob. 1PECh. 16.6 - Prob. 1RCCh. 16.7 - Prob. 1RCCh. 16.8 - Prob. 1PECh. 16.8 - Rank the following acids from strongest to...Ch. 16.9 - Prob. 1PECh. 16.9 - Practice Exercise Predict whether the following...Ch. 16.9 - Prob. 1RCCh. 16.10 - Prob. 1RCCh. 16.11 - Prob. 1PECh. 16.11 - Prob. 1RCCh. 16 - Prob. 16.1QPCh. 16 - Prob. 16.2QPCh. 16 - Prob. 16.3QPCh. 16 - Prob. 16.4QPCh. 16 - Prob. 16.5QPCh. 16 - Prob. 16.6QPCh. 16 - Prob. 16.7QPCh. 16 - Prob. 16.8QPCh. 16 - Prob. 16.9QPCh. 16 - Prob. 16.10QPCh. 16 - Prob. 16.12QPCh. 16 - Prob. 16.13QPCh. 16 - Prob. 16.14QPCh. 16 - 16.15 Calculate the hydrogen ion concentration for...Ch. 16 - 16.16 Calculate the hydrogen ion concentration in...Ch. 16 - Prob. 16.17QPCh. 16 - Prob. 16.18QPCh. 16 - 16.19 Complete this table for a...Ch. 16 - Prob. 16.20QPCh. 16 - Prob. 16.21QPCh. 16 - Prob. 16.22QPCh. 16 - Prob. 16.23QPCh. 16 - Prob. 16.24QPCh. 16 - Prob. 16.25QPCh. 16 - Prob. 16.26QPCh. 16 - Prob. 16.27QPCh. 16 - Prob. 16.28QPCh. 16 - Prob. 16.29QPCh. 16 - Prob. 16.30QPCh. 16 - Prob. 16.31QPCh. 16 - Prob. 16.32QPCh. 16 - Prob. 16.33QPCh. 16 - Prob. 16.34QPCh. 16 - Prob. 16.35QPCh. 16 - Prob. 16.36QPCh. 16 - Prob. 16.37QPCh. 16 - Prob. 16.38QPCh. 16 - Prob. 16.39QPCh. 16 - 16.40 Which of the following solutions has the...Ch. 16 - Prob. 16.41QPCh. 16 - Prob. 16.42QPCh. 16 - Prob. 16.43QPCh. 16 - Prob. 16.44QPCh. 16 - Prob. 16.45QPCh. 16 - Prob. 16.46QPCh. 16 - 16.47 A 0.040 M solution of a monoprotic acid is...Ch. 16 - Prob. 16.48QPCh. 16 - Prob. 16.49QPCh. 16 - 16.50 Write all the species (except water) that...Ch. 16 - Prob. 16.51QPCh. 16 - Prob. 16.52QPCh. 16 - Prob. 16.53QPCh. 16 - Prob. 16.54QPCh. 16 - Prob. 16.55QPCh. 16 - Prob. 16.56QPCh. 16 - 16.57 What is the original molarity of a solution...Ch. 16 - Prob. 16.58QPCh. 16 - Prob. 16.59QPCh. 16 - Prob. 16.60QPCh. 16 - Prob. 16.61QPCh. 16 - Prob. 16.62QPCh. 16 - Prob. 16.63QPCh. 16 - Prob. 16.64QPCh. 16 - Prob. 16.65QPCh. 16 - Prob. 16.66QPCh. 16 - Prob. 16.67QPCh. 16 - Prob. 16.68QPCh. 16 - Prob. 16.69QPCh. 16 - Prob. 16.70QPCh. 16 - Prob. 16.71QPCh. 16 - Prob. 16.72QPCh. 16 - Prob. 16.73QPCh. 16 - Prob. 16.74QPCh. 16 - Prob. 16.75QPCh. 16 - Prob. 16.76QPCh. 16 - Prob. 16.77QPCh. 16 - Prob. 16.78QPCh. 16 - Prob. 16.79QPCh. 16 - Prob. 16.80QPCh. 16 - Prob. 16.81QPCh. 16 - Prob. 16.82QPCh. 16 - Prob. 16.83QPCh. 16 - Prob. 16.84QPCh. 16 - Prob. 16.85QPCh. 16 - Prob. 16.86QPCh. 16 - Prob. 16.87QPCh. 16 - Prob. 16.88QPCh. 16 - Prob. 16.89QPCh. 16 - Prob. 16.90QPCh. 16 - Prob. 16.91QPCh. 16 - Prob. 16.92QPCh. 16 - 16.93 Most of the hydrides of Group 1A and Group...Ch. 16 - Prob. 16.94QPCh. 16 - Prob. 16.95QPCh. 16 - Prob. 16.96QPCh. 16 - Prob. 16.97QPCh. 16 - Prob. 16.98QPCh. 16 - Prob. 16.99QPCh. 16 - 16.100 Hydrocyanic acid (HCN) is a weak acid and a...Ch. 16 - Prob. 16.101QPCh. 16 - Prob. 16.102QPCh. 16 - Prob. 16.103QPCh. 16 - Prob. 16.104QPCh. 16 - 16.105 You are given two beakers containing...Ch. 16 - Prob. 16.106QPCh. 16 - Prob. 16.107QPCh. 16 - Prob. 16.108QPCh. 16 - Prob. 16.109QPCh. 16 - Prob. 16.110QPCh. 16 - Prob. 16.111QPCh. 16 - Prob. 16.112QPCh. 16 - Prob. 16.113QPCh. 16 - Prob. 16.114QPCh. 16 - Prob. 16.115QPCh. 16 - Prob. 16.116QPCh. 16 - Prob. 16.117QPCh. 16 - Prob. 16.118QPCh. 16 - Prob. 16.119QPCh. 16 - Prob. 16.120QPCh. 16 - Prob. 16.121SPCh. 16 - Prob. 16.122SPCh. 16 - Prob. 16.123SPCh. 16 - Prob. 16.124SPCh. 16 - Prob. 16.125SPCh. 16 - Prob. 16.126SPCh. 16 - Prob. 16.127SPCh. 16 - Prob. 16.128SPCh. 16 - Prob. 16.129SPCh. 16 - 16.130 Use the data in Appendix 2 to calculate the...
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