Answer to Question #131060 in Chemistry for davey

Question

How many millilitres of 6.0 M HNO3 would you add
to 300 mL of a buffer solution that is 0.250 M
CH3COOH and 0.560 M NaCH3COO to change the
pH from 5.09 to 5.03? Ka of acetic acid, CH3COOH,
is 1.8 × 10-5
.

Answer 1

The buffer action of the solution is based on the dissociation reaction of acetic acid.

CH₃COOH $\rightarrow$ CH₃COO^- + H⁺.

Upon addition of HNO₃, the equilibrium will shift to the left, increasing the concentration of the neutral form and decreasing the concentration of the anionic form. Assuming the complete reaction of the added hydrogen ions with the acetate anion, and since the final pH falls within one pH unit of the pK_a of acetic acid, we can use Henderson-Hasselbalch equation and the new pH:

$5.03 = 4.74 + \text{log}\frac{[CH_3COO^-]}{[CH_3COOH]}$

$1.93 = \frac{0.560 - V·6/300}{0.250+V·6/300}$

$V = 1.3$ mL.

Answer: 1.3 mL of 6.0 M HNO₃