Please just answer part (C) because A and B have been already answered correclty
ID: 794915 • Letter: P
Question
Please just answer part (C) because A and B have been already answered correclty so I just need C. Also, no need for steps, just the final correct answer
A 25.0-mL sample of a 0.110 M solution of aqueous trimethylamine is titrated with a 0.138 M solution of HCI. Calculate the pH of the solution after 10.0, 20.0, and 30.0 mL of acid have been added; pKb of (CH3)3N = 4.19at 25degreeC. pH after 10.0 mL of acid have been added: pH after 20.0 mL of acid have been added: pH after 30.0 mL of acid have been added:Explanation / Answer
The answer is: pH = 1.60
(CH3)3N + HCl => (CH3)3NH+ + Cl-
Initial moles of (CH3)3N = 25.0/1000 x 0.110 = 0.00275 mol
Moles of HCl added = 30.0/1000 x 0.138 = 0.00414 mol
Excess moles of HCl = 0.00414 - 0.00275 = 0.00139 mol
Total volume = 25.0 + 30.0 = 55.0 mL = 0.055 L
[H+] = excess moles of HCl/total volume
= 0.00139/0.055 = 0.02527
pH = -log[H+] = -log(0.02527) = 1.60