Chapter 16, Problem 60PS

### 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

# A solution of hydrofluoric acid, HF, has a pH of 2.30. Calculate the equilibrium concentration of HF, F‒, and H 3 O + and calculate the amount of HF originally dissolved per litre.

Interpretation Introduction

Interpretation:

The equilibrium concentration of HF, F-, and H3O+ has to be determined and the amount of HF originally dissolved per litre 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.

HA + H2OH3O++ A-Ka[H3O+][A-][HA]

Ka is an acid constant for equilibrium reactions.

BOH + H2OB++ OH-Ka[B+][OH-][BOH]

Kb is a base constant for equilibrium reaction.

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

Explanation

Given pH = 2.30

pHÂ =Â -log[H3O+]2.30Â =Â -log[H3O+][H3O+]Â =Â 0.005M

HF reacts with water. The equilibrium chemical reaction is as follows.

HF(aq)Â +Â H2O(aq)Â â‡ŒF-(aq)Â +Â H3O+(aq)

The equilibrium expression:

KaÂ =Â [F-][H3O+][HF]

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 HF(aq)Â +Â H2O(l)Â â‡ŒF-(aq)Â +Â H3O+(aq)Â Â Â IÂ Â Â Â Â Â Â Â Â xÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â --Â Â Â Â Â Â Â Â Â Â Â --Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â --CÂ Â Â Â Â Â -0.005Â Â Â Â Â Â Â Â Â --Â Â Â Â Â Â Â Â Â +0.005Â Â Â Â Â Â Â Â Â Â +0.005EÂ Â Â Â (Â xÂ -0.005)Â Â Â Â Â --Â Â Â Â Â Â Â Â Â Â 0.005Â Â Â Â Â Â Â Â Â Â Â Â Â 0.005

Ka of HF is 7

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