Question

Calculate the pH after 0.15 mole of NaOH is added to 1.07 L of a solution that is 0.59 M HNO2 and 1.05 M KNO2, and calculate the pH after 0.30 mole of HCl is added to 1.07 L of the same solution of HNO2 and KNO2.

0.15 mole of NaOH __________

0.30 mole of HCl __________

Answer 1

First calculate no of moles of HNO₂ and KNO₂ in buffer solution

We have , Molarity = No of moles of solute / Volume of solution in L

Therefore, No of moles of Solute = Molarity x Volume of solution in L

No of moles of HNO₂= 0.59 mol / L x 1.07 L

= 0.63 mol

No of moles of KNO₂ = 1.05 mol / L x 1..07 L

= 1.12 mol

pH of buffer after 0.15 mol NaOH

When NaOH is added to a buffer it will react with acidic component of buffer namely HNO₂ and produces NO₂^-base  . Due to addition of NaOH , [ HNO₂ ] decreases while that of [NO₂^-] increases. Hence new concentration will be

Concentration	HNO2	NaOH	NO2-
Initial	0.63	0.15	1.12
Change	- 0.15	-0.15	+0.15
Equilibrium	0.48	0.00	1.27

We know that, pH of buffer is calculated by using Henderson's equation

pH = pKa + log [ salt] / [Acid]

Here pKa of HNO₂ = 3.15 , [ salt] = [ NO₂^-] = 1.27 and [Acid] = [ HNO₂ ] = 0.48

Therefore, pH = 3.15 + log 1.27 / 0.48 ( Here volume is not considered as it is same for both acid and salt)

= 3.15 + 0.42

= 3.57

ANSWER: pH of buffer after addition of 0.15 mol NaOH = 3.57

When HCl is added to a buffer it will react with basic component of buffer namely NO₂^- and produces HNO₂  acid . Due to addition of HCl , [ HNO₂ ] increases while that of [NO₂^-] decreases. Hence new concentration will be

Concentration	HNO2	HCl	NO2-
Initial	0.63	0.30	1.12
Change	+ 0.30	-0.30	-0.30
Equilibrium	0.93	0.00	0.82

We know that, pH of buffer is calculated by using Henderson's equation

pH = pKa + log [ salt] / [Acid]

Here pKa of HNO₂ = 3.15 , [ salt] = [ NO₂^-] = 0.82 and [Acid] = [ HNO₂ ] = 0.93

Therefore, pH = 3.15 + log 0.82 / 0.93 ( Here volume is not considered as it is same for both acid and salt)

= 3.15 - 0.055

= 3.10

ANSWER: pH of buffer after addition of 0.30 mol HCl = 3.10