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Chemistry & Chemical Reactivity

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

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BuyFindarrow_forward

Chemistry & Chemical Reactivity

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

What are the equilibrium concentrations of HF, F ion, and H3O+ ion in a 0.00150 M solution of HF? What is the pH of the solution?

Interpretation Introduction

Interpretation:

Equilibrium concentration of HF and F- and the pH of a 0.00150 M solution of HF is to be determined

Concept introduction:

The pH of a solution of acetic acid can be calculated by using the hydronium ion concentration by using the expression, pH=log[H3O+].

Hydronium ion concentration is calculated by considering the equilibrium conditions and from the value of acid dissociation constant.

The concentration of acetic acid and acetate ion is determined by using equilibrium condition The concentration of acetate ion and hydronium ion is equal at equilibrium according to the reaction stoichiometry.

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 a base constant for equilibrium reaction.

BOH + H2OB++ OH-Ka[B+][OH-][BOH]

Explanation

The equilibrium reaction is as follows.

HF + H2H3O+ + F-

The constant for the reaction is as follows.

Ka=  [H3O+][F][HF]

For this reaction enter the concentrations before equilibrium established, the change that occurs as the reaction proceeds to equilibrium and the concentrations hen equilibrium has been established.

                  HF(aq) +          H2               F  +                  H3O+             0.00150                                    0                      0C-x+x+xE(0.00150 - x)xx

Substitute the all value from the ICE table in constant.

From ICE table the expression for Ka;

Ka=[F][H3O+][HF]

Substituting equilibrium concentrations for acetate ion, hydronium ion, and HF;

Ka=x2(0.0015x)

Substituting the value of acid dissociation constant (Ka) in the above equation,

7.2×104=x2(0

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