# Nitric acid, HNO2 is a weak acid with Ka= 4.5 x 10-4. It dissociates according to: HNO2 (aq) H+ (aq) + NO2- (aq)

**Discipline:** Chemistry

**Type of Paper:** Question-Answer

**Academic Level:** Undergrad. (yrs 3-4)

**Paper Format:** APA

**Pages:**1

**Words:**275

Question

(6)

HNO2 (aq) <--> H+ (aq) + NO2- (aq)

a.) What is the pH of a buffer solution which is 0.80 M in NO2- and 0.40 M in HNO2?

b.) What will be the pH of the resulting solution when 2.0 mL of 0.5 M HCl are added to 50 mL of this (from part a) buffer solution?

c.) What will be the pH of the resulting solution when 2.0 mL of 0.5 M NaOH are added to 50 mL of this (from part a) buffer solution?

An answer accompanied by a complete, step by step solution would be greatly appreciated. Thanks

**Expert Answer**

(a)

Consider the dissociation reaction of nitrous acid as shown below:

Write the Henderson-Hasselbalch equation for the given reaction as shown below:

Consider the given data:

Substitute the values in above equation as shown below:

**The pH of the buffer solution is 3.65.**

is the weak acid and is the conjugate base of the weak acid. The concentration of is given as 0.40 M and concentration of is given as 0.80 M.

The dissociation constant, of is given as.

Substituting all the values in the Henderson-Hasselbalch equation and the is obtained as, 3.65.

**(b)**

Consider the dissociation reaction of nitrous acid as shown below:

Consider the data shown below:

Concentration of HCl is given as 0.5 M, concentration of is given as 0.40 M and concentration of is given as 0.80 M. Volume of is 0.2 mL and volume of the buffer solution is 50 mL.

Moles of each species are found by multiplying molarity with volume of the solution in liters. Therefore, moles of , and HCl is found to be 0.04, 0.02 and 0.001 moles respectively.

Consider the ICE table for the change in the concentration of each species after addition of acid:

The first row is the initial moles.

The initial moles of is 0.001, 0.04 and 0.02 moles respectively.

The second row is the change in moles. Assuming the net direction of the reaction is towards the product, which gives a negative sign for reactants and positive sign for products. Let the change be . Then the change in moles of and is mol and the change in moles of is mol.

The third row is the equilibrium moles. It is just the initial moles added to the change in moles. At equilibrium, the moles of is mol, moles of is mol and moles of is mol.

Consider the change in concentration by substituting the value of variable as shown below:

As the moles of is mol. Hence, is mol. Therefore, replacing the variable with ; the equilibrium concentration of is approximately zero, the equilibrium concentration of is 0.039 mol and the equilibrium concentration of is 0.021 mol.

Calculate the concentration of each species as shown below:

**The pH of the resulting solution is 3.62.**

Concentration of the species is found by dividing the number of moles with the volume of the solution in liters. Therefore, the concentration of and is found to be 0.78 M and 0.42 M respectively.

Substituting the value in the buffer equation, the pH is found to be 3.62.

**(c)**

Consider the following data:

The concentration of is given as 0.5M, concentration of is given as 0.40M and concentration of is given as 0.80M. Volume of is 0.2mL and volume of the buffer solution is 50mL.

Moles of each species are found by multiplying the molarity with volume of the solution in liters. Therefore, moles of, and is found to be 0.04, 0.02 and 0.001 moles respectively.

Consider the following ICE table:

The first row is the initial moles. The initial moles of is 0.001, 0.04 and 0.02 moles respectively.

The second row is the change in moles. Assuming the net direction of the reaction is towards the reactant, which gives a negative sign for product and positive sign for reactants. Let the change be . Then, the change in moles of and is mol and the change in moles of is mol.

The third row is the equilibrium moles. It is just the initial moles plus change in moles. At equilibrium, the moles of is mol, moles of is mol and moles of is mol.

Substitute the value of , i.e. moles of hydroxide and redraw the ICE table as shown below:

We know that, moles of is 0.001mol. Hence, is 0.001mol. Therefore, replacing the variable with 0.001; the equilibrium concentration of is approximately zero, the equilibrium concentration of is 0.041mol and the equilibrium concentration of is 0.019 mol.

Consider the following calculations:

**The****of the buffer solution is 3.68.**

The concentration of the species is found by dividing the number of moles with the volume of the solution in liters. Therefore, the concentration of and is found to be 0.7885 M and 0.3654 M respectively.

Substituting the value in the buffer equation, theis found to be 3.68.