Chapter 16, Problem 106GQ

### Chemistry & Chemical Reactivity

10th Edition
John C. Kotz + 3 others
ISBN: 9781337399074

Chapter
Section

### Chemistry & Chemical Reactivity

10th Edition
John C. Kotz + 3 others
ISBN: 9781337399074
Textbook Problem

# Calculate the pH of the solution that results from mixing 25.0 mL of 0.14 M formic acid and 50.0 mL of 0.070 M sodium hydroxide.

Interpretation Introduction

Interpretation:

To determine the pH of the solution.

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 an 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]

Ion product constant for water

Kw= [H3O+][OH-]       =1.00×10-14 pH = -log[H3O+]pOH= -log[OH-]

Relation between pH and pOH

pH + pOH =14

Explanation

The equilibrium chemical reaction between HCO2H and NaOH is s follows.

HCO2H(aq)+OH-(aq)â‡ŒHCO2-(aq)+H2O(aq)

Consumed amount of HCO2H:

Letâ€™s calculate the consumed moles of each one.

(25.0mlÃ—1L1000mL)Ã—(0.14molL)=Â 0.0035mol

Consumed moles of NaOH

(50.0mLÃ—1L1000mL)Ã—(0.07molL)=Â 0.0035mol

Amount of HCO2- produced upon the complete of the reaction.

=Â 0.0035Â molÂ HCO2HÂ Ã—Â 1Â molÂ HCO2-1Â molÂ HCO2H=Â 0.0035Â molÂ HCO2-

Letâ€™s calculate the volume of solutions:

TotalÂ volumeÂ ofÂ solution:=(25+50)mL=75mL=75mLÃ—1L1000mL=0.075L

ConcentrationÂ ofÂ HCO2-Â =0.0035Â mol0.075Â L=Â 0.047Â MÂ HCO2-

The equilibrium reaction of HCO2-Â andÂ H2O

HCO2-(aq)+H2Oâ‡ŒHCO2H(aq)+OH-(aq)

Kb=[HCO2H][OH-][HCO2-]KbofÂ HCO2-Â isÂ 5.6Ã—10-11

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.

Â Â Â Â Â Â Â Â Â Â HCO2-Â (aq)Â +Â H2O(l)Â â‡ŒHCO2HÂ (aq)Â +Â OH-(aq)IÂ Â Â Â Â Â Â Â Â 0

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