Question

An analytical chemist is titrating 220.3 ml of 0.36 M solution of ethylamine with a 0.26 M HNO3. Calculate the pH of the base solution after the chemist has added 125 ml of the HNO3 solution to it.

Answer 1

Answer: The pH of resulting solution is 11

Explanation:

To calculate the number of moles for given molarity, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}\times 1000}{\text{Volume of solution (in mL)}}$

• For ethylamine:

Molarity of ethylamine solution = 0.36 M

Volume of solution = 220.3 mL

Putting values in above equation, we get:

$0.36M=\frac{\text{Moles of ethylamine}\times 1000}{220.3mL}\\\\\text{Moles of ethylamine}=0.079mol$

• For Nitric acid:

Molarity of nitric acid solution = 0.26 M

Volume of solution = 125 mL

Putting values in above equation, we get:

$0.25M=\frac{\text{Moles of nitric acid}\times 1000}{125mL}\\\\\text{Moles of nitric acid}=0.031mol$

The chemical reaction for ethylamine and nitric acid follows the equation:

                   $C_2H_5NH_2+HNO_3\rightarrow C_2H_5NH_3^+NO_3^-$

Initial:          0.079          0.031

Final:             0.048          -                     0.031            

Volume of solution = 220.3 + 125 = 345.3 mL = 0.3453 L    (Conversion factor:  1 L = 1000 mL)

• To calculate the pOH of basic buffer, we use the equation given by Henderson Hasselbalch:

$pOH=pK_b+\log(\frac{[salt]}{[base]})$

$pOH=pK_b+\log(\frac{[C_2H_5NH_3^+NO_3^-]}{[C_2H_5NH_2]})$

We are given:

$pK_b$ = negative logarithm of base dissociation constant of ethylamine = 3.19

$[C_2H_5NH_3^+NO_3^-]=\frac{0.031}{0.3453}$

$[C_2H_5NH_2]=\frac{0.048}{0.3453}$

pOH = ?

Putting values in above equation, we get:

$pOH=3.19+\log(\frac{0.031/0.3453}{0.048/0.3453})\\\\pOH=3.00$

To calculate pH of the solution, we use the equation:

$pH+pOH=14\\pH=14-3.00=11$

Hence, the pH of the solution is 11