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Chemistry

9th Edition
Steven S. Zumdahl
ISBN: 9781133611097

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Chemistry

9th Edition
Steven S. Zumdahl
ISBN: 9781133611097
Textbook Problem

A sample of a certain monoprotic weak acid was dissolved in water and titrated with 0.125 M NaOH, requiring 16.00 mL to reach the equivalence point. During the titration, the pH after adding 2.00 mL NaOH was 6.912. Calculate Ka for the weak acid.

Interpretation Introduction

Interpretation:

A sample of certain monoprotic weak dissolved in water and titrated with 0.125M NaOH is given. The pH value after the addition of a given amount of NaOH is given. The value of Ka for the weak acid is to be calculated.

Concept introduction:

A solution that resists a change in the pH on addition of an acid or an alkali is termed as a buffer solution.

Explanation

Explanation

To find the number of moles of monoprotic acid at the equivalence point

Given

The volume of NaOH is 16.00mL(0.016L) .

The molarity of NaOH is 0.125M .

The moles of NaOH that will be present at the equivalence point is calculated by the formula,

Numberofmoles=Volume(L)×Molarity

Substitute the value of volume and molarity of NaOH in the above expression.

Numberofmoles=0.016L×0.125M=2.0×10-3mol_

Therefore, the number of moles of the monoprotic acid initially present, that is equal to the number of moles of NaOH at the equivalence point, is 2.0×10-3mol_ .

To find the moles of monoprotic acid remaining

The moles of NaOH that will be present in the given 2.0mL of 0.125M solution is calculated by the formula,

Numberofmoles=Volume(L)×Molarity

Substitute the value of volume and molarity of NaOH in the above expression.

Numberofmoles=0.002L×0.125M=2.5×104mol

Therefore, the moles of the acid that will be consumed during titration is 2.5×104mol .

The moles of the acid remaining in the solution is calculated by the formula,

Molesofacidremaining=InitialmolesofacidMolesofacidthatreacted

Substitute the value of the initial moles of the acid and the moles of the acid that get consumed in the above expression.

Molesofacidremaining=(2.0×103mol)(2

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