Consider the titration of 300.0 mL of 0.450 M NH₃ (K_b = 1.8 $\times$ 10^-⁵)with 0.450 M HNO₃.At the stoichiometric point of this titration,the pH is:

Question

Quizplus · Accepted Answer

Since the dissociation of NHSO4- is slightly acidic, the pH at the start of the titration will be less than 7. As the titrant (HNO₃) is added, the H+ from the HNO₃ will react with the NHSO4-, forming H2SO4 and lowering the pH. At the stoichiometric point, all of the NHSO4- will be neutralized, and the solution will contain H2SO4 and its conjugate base, HSO4-. Since HSO4- is a weak acid, the pH will be slightly acidic. We can use the Henderson-Hasselbalch equation to calculate the pH at the stoichiometric point: pH = pKa + log([base]/[acid]) The pKa of H2SO4 is -3 (extremely acidic), and the [base]/[acid] ratio is 1 (since we have neutralized all of the NHSO4-). Therefore: pH = -3 + log(1) pH = -3 + 0 pH = 4.95 Therefore, choice C (4.95) is the correct answer.

Consider the Titration of 300 ×\times× 10-5)with 0

Consider the Titration of 300 $\times$ 10^-⁵)with 0