Question

When a 26.3 mL sample of a 0.465 M aqueous nitrous acid solution is titrated with a 0.461 M aqueous barium hydroxide solution, what is the pH after 19.9 mL of barium hydroxide have been added

Answer 1

Answer: The pH of the solution after addition of barium hydroxide is 3.78

Explanation:

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

$\text{Molarity of the solution}=\frac{\text{Moles of solute}\times 1000}{\text{Volume of solution (in mL)}}$      .......(1)

• For nitrous acid:

Molarity of nitrous acid = 0.465 M  

Volume of solution = 26.3 mL

Putting values in equation 1, we get:

$0.465M=\frac{\text{Moles of nitrous acid}\times 1000}{26.3mL}\\\\\text{Moles of nitrous acid}=\frac{0.465\times 26.3}{1000}=0.0122mol$

• For barium hydroxide:

Molarity of  barium hydroxide = 0.461 M  

Volume of solution = 19.9 mL

Putting values in equation 1, we get:

$0.461M=\frac{\text{Moles of  barium hydroxide}\times 1000}{19.9mL}\\\\\text{Moles of  barium hydroxide}=\frac{0.461\times 19.9}{1000}=0.0092mol$

The chemical reaction for nitrous acid and barium hydroxide follows the equation:

                $2HNO_2+Ba(OH)_2\rightarrow Ba(NO_2)_2+2H_2O$  

Initial:       0.0122    0.0092          

Final:         0.003          -                0.0092

Volume of solution = 26.3+ 19.9 = 46.2 mL = 0.0462 L   (Conversion factor:  1 L = 1000 mL)

To calculate the pH of acidic buffer, we use the equation given by Henderson Hasselbalch:

$pH=pK_a+\log(\frac{[salt]}{[acid]})$

$pH=pK_a+\log(\frac{[NO_2^-]}{[HNO_2]}$

We are given:

$pK_a$ = negative logarithm of acid dissociation constant of nitrous acid = 3.29

$[NO_2^-]=\frac{0.0092}{0.0462}$  

$[HNO_2]=\frac{0.003}{0.0462}$

pH = ?

Putting values in above equation, we get:

$pH=3.29+\log(\frac{0.0092/0.0462}{0.003/0.0462})\\\\pH=3.78$

Hence, the pH of the solution after addition of barium hydroxide is 3.78