   Chapter 15, Problem 121IP

Chapter
Section
Textbook Problem

# Calculate the pH of a solution prepared by mixing 250. mL of 0.174 m aqueous HF (density = 1.10 g/mL) with 38.7 g of an aqueous solution that is 1.50% NaOH by mass (density = 1.02 g/mL). (Ka for HF = 7.2 × 10−4.)

Interpretation Introduction

Interpretation:

The titration of NH3 with HCl is given. The pH at which equivalence point occurs is to be calculated.

Concept introduction:

The solution of weak acid and their conjugate base which upon addition of an acid or base resists the change in pH is called buffer solution. The pH of the buffer solution remains constant.

To determine: The pH of given solution formed by mixing HF with NaOH .

Explanation

Explanation

Given

The molality of HF is 0.174m .

The volume of HF is 25.0mL .

The density of HF is 1.10g/mL .

The mass of NaOH solution is 38.7g

The mass percent of NaOH is 1.50% .

The density of NaOH is 1.02g/mL .

The Ka of HF is 7.2×104

The density of any substance is given as,

Density=MassVolume (1)

Rearrange the equation to find the mass of the substance as,

Mass=Density×Volume

Substitute the value of mass and volume of HF in the above equation as,

Mass=Density×Volume=1.10gmL×250.0mL=275g

The conversion of g into kg is done as,

1g=0.001kg

Hence the conversion of 275g into kg is done as,

275g=275×0.001kg=0.275kg

Molality of a solution is given as,

Molality=NumberofmolesofsubstanceMassofsolventinkg

Rearrange the equation to obtain the number of moles as,

Numberofmolesofsubstance=Massofsolventinkg×Molality

Substitute the mass of solvent that is HF and molality of solution in the above equation to obtain the number of moles as,

Numberofmolesofsubstance=Massofsolventinkg×Molality=0.275kg×0.174molekg=0.0479mole

Rearrange the equation (1) to obtain the volume of the solution as,

Volume=MassofthesubstanceDensity

Substitute the mass of NaOH and density of NaOH solution to obtain its volume as,

Volume=MassofthesubstanceDensity=38.7g1.02gmL=37.94mL

The conversion of mL into L is done as,

1mL=0.001L

Hence the conversion of 37.94mL into L is done as,

37.94mL=37.94×0.001L=0.0379L

The amount of NaOH present in 100g of NaOH solution =1.50g

The amount of NaOH present in 1.0g of NaOH solution =1.50g100g

The amount of NaOH present in 38.75g of NaOH solution =1.50g100g×38.75g=0.5813g

The molar mass of NaOH is 40.0g/mole .

The relationship between number of moles and mass of substance is given as,

Numberofmoles=GivenmassMolarmass

Substitute the obtained mass and molar mass of substance in the above equation as,

Numberofmoles=GivenmassMolarmass=0

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