Calculate the mass of sodium hydroxide that must be added to 1.00 L of 1.00- M HC 2 H 3 O 2 to double the pH of the solution (assume that the added NaOH does not change the volume of the solution).

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Interpretation Introduction Interpretation: The mass of NaOH required to be added to 1.00 L of 1.00 M acetic acid ( CH 3 COOH ) in order to double the pH of the solution is to be calculated. Concept introduction: Weak acid dissociates up to a very small extent in an aqueous solution and on addition of a base, neutralization reaction occurs and formation of a salt takes place. This salt acts as a buffer solution. To determine: The mass of NaOH required to be added to 1.00 L of 1.00 M acetic acid ( CH 3 COOH ) in order to double the pH of the solution. Explanation Explanation The pH of given solution of acetic acid is 2 . 3 8 _ . A weak acid like acetic acid dissociates partially in aqueous solution and ICE (Initial Change Equilibrium) is used to calculate the concentration of products at equilibrium. CH 3 COOH + H 2 O ⇌ H 3 O + + CH 3 COO − Initial ( M ) : 1 M 0 0 Change ( M ) : − x x x Equilibrium ( M ) : 1 − x x x The acid dissociation constant formula for the above reaction is. K a = [ H 3 O + ] [ CH 3 COO − ] [ CH 3 COOOH ] Substitute the values of concentration and standard value of K a in above equation. 1.8 × 10 − 5 = x × x 1.00 − x x 2 + 1.8 × 10 − 5 − 1.8 × 10 − 5 = 0 x = 0 .0042 M Therefore, the [ H + ] is 0.0042 M . The pH is calculated by the formula. pH = − log [ H + ] Substitute the value of [ H + ] in the above expression. pH = − log ( 0.0042 ) pH = 2 . 3 8 _ The mass of NaOH added is 2 0 g _ . The addition of NaOH causes the neutralization reaction and the resulting product acts as a buffer solution. The amount of NaOH added is assumed to be x M then the amount of salt formed is x M after reacting with x M of acid

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Chemistry: An Atoms First Approach

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Steven S. Zumdahl + 1 other

Publisher: Cengage Learning

ISBN: 9781305079243

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Interpretation Introduction

Interpretation: The mass of NaOH required to be added to 1.00L of 1.00 M acetic acid (CH3COOH) in order to double the pH of the solution is to be calculated.

Concept introduction: Weak acid dissociates up to a very small extent in an aqueous solution and on addition of a base, neutralization reaction occurs and formation of a salt takes place. This salt acts as a buffer solution.

To determine: The mass of NaOH required to be added to 1.00L of 1.00 M acetic acid (CH3COOH) in order to double the pH of the solution.

Explanation of Solution

Explanation

The pH of given solution of acetic acid is 2.38_.

A weak acid like acetic acid dissociates partially in aqueous solution and ICE (Initial Change Equilibrium) is used to calculate the concentration of products at equilibrium.

CH3COOH+H2O⇌H3O++CH3COO−Initial(M) :1 M00Change(M) :−xxxEquilibrium(M):1−xxx

The acid dissociation constant formula for the above reaction is.

Ka=[H3O+][CH3COO−][CH3COOOH]

Substitute the values of concentration and standard value of Ka in above equation.

1.8×10−5=x×x1.00−xx2+1.8×10−5−1.8×10−5=0x=0.0042 M

Therefore, the [H+] is 0.0042 M.

The pH is calculated by the formula.

pH=−log[H+]

Substitute the value of [H+] in the above expression.

pH=−log(0.0042)pH=2.38_

The mass of NaOH added is 20 g_.

The addition of NaOH causes the neutralization reaction and the resulting product acts as a buffer solution. The amount of NaOH added is assumed to be x M then the amount of salt formed is x M after reacting with x M of acid

Calculate the mass of sodium hydroxide that must be added to 1.00 L of 1.00- M HC 2 H 3 O 2 to double the pH of the solution (assume that the added NaOH does not change the volume of the solution).

Chemistry: An Atoms First Approach

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Chemistry: An Atoms First Approach

Calculate the mass of sodium hydroxide that must be added to 1.00 L of 1.00-M HC₂H₃O₂ to double the pH of the solution (assume that the added NaOH does not change the volume of the solution).

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Calculate the mass of sodium hydroxide that must be added to 1.00 L of 1.00- M HC 2 H 3 O 2 to double the pH of the solution (assume that the added NaOH does not change the volume of the solution).

Chemistry: An Atoms First Approach

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Chemistry: An Atoms First Approach