Calculate the pH after 0.10 mole of NaOH is added to 1.00 L of the solution in Exercise 31, and calculate the pH after 0.20 mole of HCl is added to 1.00 L of the solution in Exercise 31.

Interpretation:

The pH value after 0.10 mol NaOH is added to 1 L of the solution in Exercise 31 and the pH value after 0.20 mol HCl is added to 1 L of the solution in exercise 31 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. The pH is calculated by the formula,

pH=−log10[H+]

To determine: The pH value after 0.10 mol NaOH is added to 1 L of the solution in exercise 31 ; the pH value after 0.20 mol HCl is added to 1 L of the solution in exercise 31 .

Explanation

Given

Volume of solution is 1.0 L .

Number of moles of NaOH added is 0.10 mol .

Refer Exercise 31 .

The pH value of the solution of 1.00 M HNO2 and 1.00 M NaNO2 is 3.40 .

The concentration of HNO2 is 1.00 M .

The concentration of NaNO2 is 1.00 M .

Formula

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

Moles of compound=Concentration of compound×Volume of solution (1)

Substitute the values of concentration and volume of HNO2 in the above equation.

Moles of compound=Concentration of compound×Volume of solution=1.0 M×1.0 L=1.0 mol_

Substitute the values of concentration and volume of NaNO2 in equation (1).

Moles of compound=Concentration of compound×Volume of solution=1.0 M×1.0 L=1.0 mol_

To determine the final number of moles of HNO2 and NaNO2

The dissociation reaction of HNO2 is,

HNO2+OH−→NO2−+H2O

Initial moles of HNO2 is 1.00 mol .

On addition of the given concentration of NaOH to the buffer solution, the OH− ions react with HNO2 and lead to the formation of NO2− ions.

The final moles of NO2− are,

1.00 mol+0.10 mol=1.10 mol_

The final moles of HNO2 are,

1.00 mol−0.10 mol=0.90 mol_

To determine the value of pKa

The value of Ka of HNO2 is 4.3×10−4 .

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(4.3×10−4)=3.37_

To determine the   pH

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

Therefore, the concentration of HNO2 is 0.90 M and of NO2− is 1.10 M

The value of pKa is 3.37 .

Formula

The pH is calculated using the Henderson-Hassel Bach equation,

pH=pKa+log[NO2−][HNO2]

Where,

• pH is the measure of H+ ions.
• pKa is the measure of acidic strength.
• [NO2−] is concentration of NO2−