   Chapter 15, Problem 68E

Chapter
Section
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,

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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