   # Calculate the hydronium ion concentration and pH for a 0.015M solution of sodium formate NaHCO 3 . ### Chemistry & Chemical Reactivity

9th Edition
John C. Kotz + 3 others
Publisher: Cengage Learning
ISBN: 9781133949640

#### Solutions

Chapter
Section ### Chemistry & Chemical Reactivity

9th Edition
John C. Kotz + 3 others
Publisher: Cengage Learning
ISBN: 9781133949640
Chapter 16, Problem 60PS
Textbook Problem
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## Calculate the hydronium ion concentration and pH for a 0.015M solution of sodium formate NaHCO 3 .

Interpretation Introduction

Interpretation:

The hydronium ion concentration and pH for a 0.015M solution of sodium formate (NaHCO3) has to be calculated.

Concept introduction:

Equilibrium constants:

The equilibrium constant is used to quantitative measurement of the strength of the acid and bases in the water.

Ka is acid constant for equilibrium reactions.

HA + H2OH3O++ A-Ka[H3O+][A-][HA]

Kb is base constant for equilibrium reaction.

BOH + H2OB++ OH-Ka[B+][OH-][BOH]

Ionic product constant for water  Kw= [H3O+][OH-]       =1.00×10-14 pH = -log[H3O+]pOH= -log[OH-]Reltion between pH and pOH pH + pOH =14

### Explanation of Solution

Sodium formate (NaHCO2) is dissociates into Na+ and HCO2- ions. Sodium ions are not affecting the pH of the solution. The pH of the solution completely depending upon the HCO2- ions.

HCO2- ions react with water the chemical equilibrium equation is as follows.

HCO2-(aq) + H2O(l)HCO2H(aq) + OH-(aq)

The equilibrium expression:

Kb = [HCO2H][OH-][HCO2-]

Enter the ICE table the concentrations before equilibrium is established, the change that occurs as the reaction proceeds to equilibrium and the concentrations when equilibrium has been achieved.

HCO2-(aq) + H2O(l) HCOOH(aq) + OH-(aq)   I         0.015             --                 --                     --C         -x                  --               +x                    +x    E      (0.015-x)         --                 x                      x

Kb of  HCO2- is 5.6×10-11

Kb = [HCO2H][OH-][HCO2-]

5

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