Consider the titration of 30.0 mL of 0.050 M NH3 with 0.025 M HCl. Calculate the pH after the following volumes of titrant have been added:

Question

a. 61.0 mL
b. 65.0 mL

im having a hard time even starting this question so any help would be greatly appreciated. thank you

Answer 1

The first step is to determine where you are on the titration curve because that tells you what you have in the solution. To do that we determine where the equivalence point is located; i.e., what volume of HCl is needed to exactly neutralize the initial 15 millimoles NH3.
mL NH3 x M NH3 = mL HCl x M HCl
30 x 0.05 = mL HCl x 0.025
mL HCl = 60 mL. Let's show that below.

Here are the data:
millimoles NH3 = mL x M = 30 x 0.05 = 15 (initial)
millimoles HCl @ 61 mL = 61 x 0.025 = 15.25
millimoles HCl @ 65 mL = 65 x 0.025 = 16.25
Use these data to calculate the equivalence point as follows:
.........................NH3 + HCl ==> NH4Cl
I........................15.........0...............0
add...............................15..................
C......................-15.........-15............+15
E.........................0..........0...................15
So at the equivalence point you have only the salt, NH4Cl, in solution, with 15 mmols NH4Cl in 90 mL H2O.
Therefore, the problem is asking you to determine the pH of the solution AFTER the equivalence point where you have the salt NH4Cl, an excess of HCl now, and water to dilute everything.
How much NH3 did you start with? That's 15 mmols and 30 mL
How much HCl has been added @ 65 mL? That's 16.25 mmols and 65 mL.
How much HCl is in excess? That's 16.25 - 15 = 1.25 mmols HCl in 95 mL.
(HCl) = 1.25 mmols/95 mL = ?. Convert that to pH = -log(HCl)
The 61 mL is done the same way.
Post your work if you get stuck.