   Chapter 16, Problem 106GQ

Chapter
Section
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)HCO2-(aq)+H2O(aq)

Consumed amount of HCO2H:

Let’s calculate the consumed moles of each one.

(25.0ml×1L1000mL)×(0.14molL)= 0.0035mol

Consumed moles of NaOH

(50.0mL×1L1000mL)×(0.07molL)= 0.0035mol

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

= 0.0035 mol HCO2H × 1 mol HCO2-1 mol HCO2H= 0.0035 mol HCO2-

Let’s calculate the volume of solutions:

Total volume of solution:=(25+50)mL=75mL=75mL×1L1000mL=0.075L

Concentration of HCO2- =0.0035 mol0.075 L= 0.047 M HCO2-

The equilibrium reaction of HCO2- and H2O

HCO2-(aq)+H2OHCO2H(aq)+OH-(aq)

Kb=[HCO2H][OH-][HCO2-]Kbof HCO2- is 5.6×10-11

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.

HCO2- (aq) + H2O(l) HCO2H (aq) + OH-(aq)I         0

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