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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

Calculate the pH after 0.10 mole of NaOH is added to 1.00 L of the solution in Exercise 32, and calculate the pH after 0.20 mole of HCl is added to 1.00 L of the solution in Exercise 32.

Interpretation Introduction

Interpretation:

The pH value after 0.10mol NaOH is added to 1.0L of the solution in Exercise 32 and the pH value after 0.20mol HCl is added to 1.0L of the solution in Exercise 32 is to be calculated.

Concept introduction:

A solution that contains a mixture of a weak acid and its conjugate base is known as a buffer solution.

The pH value is the measure of H+ ions. It is calculated by the formula,

pH=log10[H+]

To determine: The pH value after 0.10mol NaOH is added to 1.0L of the solution in Exercise 32 and the pH value after 0.20mol HCl is added to 1.0L of the solution in Exercise 32 .

Explanation

Explanation

To determine the number of moles of HF and KF .

Given

Volume of solution is 1.0L .

Moles of NaOH added is 0.10mol .

Refer Exercise 32 .

The pH value of the solution of 0.60M HF and 1.00M KF is 3.36 .

The concentration of HF is 0.60M .

The concentration of KF is 1.00M .

Formula

The number of moles of compound in a solution is calculated by the formula,

Molesofcompound=Concentrationofcompound×Volumeofsolution (1)

Substitute the values of concentration and volume of KF in the above equation.

Molesofcompound=Concentrationofcompound×Volumeofsolution=1.0M×1.0L=1.0mol_

Substitute the values of concentration and volume of HF in equation (1).

Molesofcompound=Concentrationofcompound×Volumeofsolution=0.6M×1.0L=0.6mol_

To determine the moles of HF and KF after addition of NaOH

Initial moles of HF is 0.6mol .

The OH from NaOH reacts with HF to give F ions. Therefore, the concentration of HF decreases by 0.1mol .

Hence, the final moles of HF is,

0.6mol0.10mol=0.50mol_

Initial moles of F is 1.00mol .

The OH from NaOH reacts with HF to give F ions. Therefore, the concentration of F increases by 0.1mol .

Hence, the final moles of F is,

1.00mol+0.10mol=1.10mol_

To determine the value of pKa

The value of Ka of HF is 7.2×104 .

The formula of pKa is,

pKa=logKa

Where,

  • Ka is acid equilibrium constant.

Substitute the value of Ka in the above equation.

pKa=logKa=log(7.2×104)=3.14_

To determine the pH value after 0.10mol NaOH is added to 1.0L of the solution of HF and KF

As the volume of solution is 1.0L , the concentration of compound is equal to number of moles.

Therefore, [HF] is 0

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