1. Calculate the pH for each of the following cases in the titration of 25.0 mL of 0.240 M pyridine, C5H5N(aq) with 0.240 M HBr(aq):
(a) before addition of any HBr
(b) after addition of 12.5 mL of HBr
(c) after addition of 22.0 mL of HBr
(d) after addition of 25.0 mL of HBr
(e) after addition of 37.0 mL of HBr
I believe I know how to do part a, I got pH=9.31. After that I don't understand what to do if HBr is added. Any help is really appreciated! Thanks!
Oh, titrations can be quite acidic, but don't worry, I'm here to help with a touch of humor!
Firstly, let's tackle the initial pH before any HBr is added. You mentioned that you got pH=9.31, which sounds about right for pyridine, a weak base. Well done!
Now, when HBr is added, it reacts with pyridine to form a salt, which then splits into its constituent ions. We'll have a buffer solution situation going on, so we won't see a drastic change in pH initially.
For part (b), start by calculating the moles of HBr added, given that its concentration is 0.240 M and the volume added is 12.5 mL. Then, use the Henderson-Hasselbalch equation to calculate the new pH. Remember, laughter is the best buffer!
For parts (c) to (e), follow the same procedure. Just keep track of the changes in volume and concentration of HBr as you go along. Take a deep breath and keep the puns flowing!
Remember, in a titration, you can usually assume that the reaction is complete and all the acid reacts with the base. With that in mind, you can calculate the pH after each addition of HBr. Don't let the acid get to you, stay calm and keep calculating!
So, give those calculations another shot, and if you get stuck, just holler and I'll be here to guide you through the chemistry circus!
To calculate the pH for each case, we need to consider the reaction that occurs between pyridine (C5H5N) and HBr:
C5H5N(aq) + HBr(aq) -> C5H5NH+(aq) + Br-(aq)
This reaction is a weak base and strong acid reaction. Pyridine is a weak base that accepts a proton from HBr to form the pyridinium ion (C5H5NH+).
(a) Before addition of any HBr:
In this case, the solution contains only pyridine. We can use the Kb value for pyridine to find the pOH and then convert it to pH.
Kb for pyridine (C5H5N): 1.7 x 10^-9
To find the pOH, we can use the equation:
pOH = -log(Kb)
pOH = -log(1.7 x 10^-9) = 8.77
To find the pH, we can use the equation:
pH = 14 - pOH
pH = 14 - 8.77 = 5.23
Therefore, the pH before addition of any HBr is 5.23.
(b) After addition of 12.5 mL of HBr:
In this case, we have added half of the initial volume of HBr to the pyridine solution. The HBr will react completely with the pyridine to form the pyridinium ion.
To calculate the concentration of pyridinium ion, we can use the equation:
nC1V1 = nC2V2
Where nC1 is the initial number of moles of pyridine, nC2 is the number of moles of pyridinium ion formed, V1 is the initial volume of pyridine, and V2 is the volume of HBr added.
nC1 = VC1
= (25.0 mL)(0.240 M)
= 6.0 mmol
nC2 = nHBr
= VHBr
= (12.5 mL)(0.240 M)
= 3.0 mmol
From the stoichiometry of the reaction, the number of moles of pyridine reacted is equal to the number of moles of pyridinium ion formed, so nC2 = 3.0 mmol.
The total volume of the solution is the sum of the initial volume of pyridine and the volume of HBr added:
Vtotal = Vpyridine + VHBr
= 25.0 mL + 12.5 mL
= 37.5 mL
To calculate the concentration of the pyridinium ion, we can use the equation:
Cpyridinium = nC2 / Vtotal
= (3.0 mmol) / (37.5 mL)
= 0.080 M
Since the reaction produces equal concentrations of pyridinium ion and Br-, the concentration of Br- is also 0.080 M.
Now we can use the Kb value for pyridinium ion (C5H5NH+) to find the pOH and then convert it to pH.
Kb for pyridinium ion (C5H5NH+): 1.7 x 10^-9
To find the pOH, we can use the equation:
pOH = -log(Kb)
pOH = -log(1.7 x 10^-9) = 8.77
To find the pH, we can use the equation:
pH = 14 - pOH
pH = 14 - 8.77 = 5.23
Therefore, the pH after addition of 12.5 mL of HBr is also 5.23.
Repeat the same calculations for parts (c) after addition of 22.0 mL of HBr, (d) after addition of 25.0 mL of HBr, and (e) after addition of 37.0 mL of HBr to find the pH in each case.
To calculate the pH at each stage of the titration, you need to consider the stoichiometry of the reaction between pyridine (C5H5N) and HBr. The reaction can be written as:
C5H5N(aq) + HBr(aq) → C5H5NH+(aq) + Br-(aq)
In the titration, HBr is the strong acid being added to the weak base pyridine. Thus, the main reaction occurring is the protonation of pyridine to form its conjugate acid, C5H5NH+.
To solve parts (b) to (e), you should follow these steps:
1. Determine the limiting reagent:
(a) The initial number of moles of pyridine in the solution can be calculated using the initial concentration and volume.
Given: Initial volume of pyridine solution = 25.0 mL = 0.025 L
Initial concentration of pyridine solution = 0.240 M
Initial moles of pyridine = initial concentration × initial volume = 0.240 M × 0.025 L
= 0.006 moles
(b) The number of moles of HBr added in each step can be calculated using the molarity and volume.
Given: Volume of HBr added = 12.5 mL, 22.0 mL, 25.0 mL, 37.0 mL (convert each to liters)
Moles of HBr added in each step = HBr concentration × volume of HBr added
2. Determine the moles of pyridine that react with the added HBr:
For each step, the moles of pyridine that react with HBr can be determined using the stoichiometry of the balanced reaction equation.
The mole ratio between pyridine and HBr is 1:1.
Therefore, the moles of pyridine reacted = moles of HBr added.
3. Calculate the final moles of pyridine:
For each step, the moles of pyridine remaining can be calculated by subtracting the moles of pyridine reacted from the initial moles of pyridine.
4. Calculate the final concentration of pyridine:
The final volume is the sum of the initial volume of pyridine solution and the volume of HBr added in each step.
The final concentration of pyridine after each step can be calculated by dividing the final moles of pyridine by the final volume.
5. Calculate the pOH and pH:
The pOH can be calculated using the concentration of pyridine.
The pH can be calculated using the pOH.
Repeat steps 1 to 5 for each step (b) to (e) to determine the pH at each stage of the titration.
Note: The pKa of pyridine is approximately 5.2, so before the addition of any HBr (part a), pyridine will act as a weak base and have a pH greater than 7.
py + HBr ==> PyH + Br^-
You know how many mols pyridine you started with. That's M x L = ?
Now take M x L of the titrant to give yo mols HBr added. Since the reaction is 1:1 you know 1 mol HBr neutralizes 1 mol pyridine.
Subtract mols py-mol HBr to see which is in excess. If py is in excess use the Henderson-Hasselbalch equation to solve for pH. If HBr is in excess, see how much in excess, determine molarity, and use pH = -log(HBr)
Those two procedures will work for all points AFTER the beginning and AFTER the equivalence point. AT THE equivalence point you must use the hydrolysis of the salt formed.
That is
.......pyH^+ + H2O==>H3O^+ + py
I.......x..............0.......0
C.......-y.............y.......y
E......x-y.............y........y
Ka for salt = (Kw/Kb for py) = (y)(y)/(x-y) and solve for y = (H3O^) and convert to pH.
Don't be confused by the two unknowns here. x is just the salt concn at the eq. pt. whatever that happens to be. I just didn't work through to see what it was.