Question

A buffer solution contains 0.240 M ammonium chloride and 0.499 M ammonia. If 0.0565 moles of perchloric acid are added to 250 mL of this buffer, what is the pH of the resulting solution? (Assume that the volume does not change upon adding perchloric acid.)

Answer 1

Answer:

The pH of the resulting solution is 9.02.

Explanation:

The initial pH of the buffer solution can be found using the Henderson-Hasselbalch equation:

$pOH = pKb + log(\frac{[NH_{4}Cl]}{[NH_{3}]})$  

$pOH = -log(1.78\cdot 10^{-5}) + log(\frac{0.240}{0.499}) = 4.43$        

$pH = 14 - pOH = 14 - 4.43 = 9.57$

Now, the perchloric acid added will react with ammonia:

$n_{NH_{3}} = 0.499moles/L*0.250 L - 0.0565 moles = 0.0683 moles$

Also, the moles of ammonium chloride will increase in the same quantity according to the following reaction:

NH₃ + H₃O⁺ ⇄ NH₄⁺ + H₂O    

$n_{NH_{4}Cl} = 0.240 moles/L*0.250L + 0.0565 moles = 0.1165 moles$

Finally, we can calculate the pH of the resulting solution:

$pOH = pKb + log(\frac{[NH_{4}Cl]}{[NH_{3}]})$  

$pOH = -log(1.78\cdot 10^{-5}) + log(\frac{0.1165 moles/0.250 L}{0.0683 moles/0.250 L}) = 4.98$  

$pH = 14 - 4.98 = 9.02$

Therefore, the pH of the resulting solution is 9.02.

I hope it helps you!