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Chemistry

9th Edition
Steven S. Zumdahl
ISBN: 9781133611097

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BuyFindarrow_forward

Chemistry

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

A student dissolves 0.0100 mole of an unknown weak base in 100.0 mL water and titrates the solution with 0.100 M HNO3. After 40.0 mL of 0.100 M HNO3 was added, the pH of the resulting solution was 8.00. Calculate the Kb value for the weak base.

Interpretation Introduction

Interpretation:

The number of moles of an unknown weak base is given. The value of Kb for the given volume of this weak base is to be calculated.

Concept introduction:

At equilibrium, the equilibrium constant expression is expressed by the formula,

Kb=ConcentrationofproductsConcentrationofreactants

To determine: The value of Kb for the given volume of the weak base.

Explanation

Explanation

To find the [OH]

The moles of HNO3 that are added are calculated by the formula,

Moles=Molarity×Volume(L)

Substitute the value of the molarity of HNO3 and the volume of the solution in the above expression.

Moles=0.1M×0.04(L)=0.004mol

The amount of base that is present after the neutralization of the acid takes place is calculated by the formula,

Excessbase=InitialmolesofbaseMolesofHNO3added

Substitute the value of the initial moles of the base and the moles of HNO3 that were added in the above expression.

Excessbase=0.010.004=0.006mol

The total volume =(100+40)mL=0.14L

The concentration of the base is calculated by the formula,

Concentration=NumberofmolesVolume(L)

Substitute the value of the number of moles and volume of the solution.

Concentration=0.006mol0.14L=0.04285M

Therefore, [B] is 0.04285M .

The reaction that occurs is,

B(Base)+H2OHB++OH

The pH is given to be 8 .

Therefore, pOH=14pH=148=6

The pOH of a solution is calculated by the formula,

pOH=log[OH]

Rearrange the above expression to calculate the value of [OH]

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