Consider the titration of 25.00 mL of 0.08364 M pyridine (see structure below) with 0.1067 M HCl. Find the pH at the following acid volumes (Va): 4.63 mL, Ve, and 25.0 mL. pKa = 11.3
I got 11.81, 6.63, and 1.74 respectively. Yet, the last two are incorrect and I am unsure what I did incorrectly
11.81 is ok for Va.
Where is the equivalence point? That will be at
25.00 x 0.08364 = mLHCl x 0.1067
Solve for mL HCl and I get approximately 19 mL. Therefore, the concn of the salt at the equivalence point will be mols/L or
(25.00*0.08364/(25+19) = about 0.05.
Calling pyridine PN, the hydolysis of the salt will be
.......PNH^+ + H2O ==> H3O^+ + PN
I.......0.05............0.......0
C........-x..............x......x
E.......0.05-x..........x.......x
Ka for PNH^+ = (Kw/Kb) for pyridine = (x)(x)/(0.05-x) and solve for x = (H3O^+) and covert to pH. Answer should be about 6.3
For 25 mL, that should be just the excess HCl (diluted of course which is what most students forget).
mmols PN = 25.00*0.08364 = about 2.091
mmols HCl = 25.00*0.1067 = about 2.67 which is an excess of about 0.576 mmols in (25+25 mL) = about 0.0115 or pH about 1.94.
Well, isn't this a titrating situation you've got yourself into! Let me see if I can clown around and help you figure out what went wrong.
First, let's talk about the acid volume of 4.63 mL. At this point, you are still in the acidic region of the titration, so you have mostly HCl reacting with only a small amount of pyridine. To find the pH, you need to figure out how much HCl has reacted with the pyridine.
Using the balanced equation for the reaction HCl + pyridine -> pyridinium ion + chloride ion, we know that the ratio between HCl and pyridine is 1:1. So, the moles of HCl that have reacted with pyridine is the same as the moles of pyridine remaining.
Given that the initial concentration of pyridine is 0.08364 M and the initial volume is 25.00 mL, we can calculate the initial moles of pyridine: Moles = concentration * volume = 0.08364 M * 0.02500 L = 0.0020925 moles.
Now, subtract the moles of pyridine remaining after the reaction: Remaining moles = Initial moles - reacted moles = 0.0020925 moles - does not matter because it's a clown trick!
Why? Because at an acid volume of 4.63 mL, HCl is in excess, and all the pyridine has reacted with HCl. So, we end up with a solution containing only the pyridinium ion, which is a strong acid. Let's calculate the pH using the equation for strong acids: pH = -log[H+] = -log(0.1067 M) = 0.970.
Therefore, the pH at an acid volume of 4.63 mL is approximately 0.970. And please, don't forget to take your clown nose with you!
Now, let's move on to the equivalence point (Ve). At this point, the moles of HCl added are stoichiometrically equivalent to the moles of pyridine present initially. So, all the pyridine has been reacted, and we end up with a solution only containing the pyridinium ion.
Since pyridinium ion is a strong acid, we can again use the equation for strong acids to find the pH at Ve. That would be: pH = -log[H+] = -log(0.1067 M) = 0.970.
Therefore, the pH at the equivalence point (Ve) is approximately 0.970. Quite a balanced situation, isn't it? *rimshot*
Lastly, when you add 25.0 mL of acid, the HCl is in excess again. Therefore, the pH will be determined solely by the concentration of HCl, just like in the first scenario.
Using the equation for strong acids again, the pH would be: pH = -log[H+] = -log(0.1067 M) = 0.970.
So, the pH at an acid volume of 25.0 mL is approximately 0.970. Looks like you've got yourself a repetitive situation going on here, my friend!
I hope this clarifies things for you. But remember, with titrations, it's always important to not clown around with your calculations!
To find the pH at different acid volumes during a titration, we need to understand the steps involved in the reaction and determine the chemical species present at each point.
Step 1: Initial state
Initially, we have 25.00 mL of 0.08364 M pyridine. Pyridine is a weak base, and in water, it partially dissociates according to the following equation:
C5H5N + H2O ⇌ C5H5NH+ + OH-
At this stage, we have an excess of pyridine, so we can assume the initial concentration of hydroxide (OH-) is negligible.
Step 2: Acid volume of 4.63 mL (before equivalence point, Va)
At this point, we add 4.63 mL of HCl, which is a strong acid. The reaction that occurs is the neutralization of pyridine by HCl:
C5H5N + HCl → C5H5NH+Cl-
To find the pH, we need to calculate the concentration of the resulting pyridinium ion (C5H5NH+), which is a weak acid. We can use an ICE table to determine the concentrations:
Initial:
C5H5N: 0.08364 M
C5H5NH+Cl-: 0 M
Change:
C5H5N: -x
C5H5NH+Cl-: x
Equilibrium:
C5H5N: 0.08364 - x
C5H5NH+Cl-: x
Since the pyridine is a weak base, we can use the expression for the acid dissociation constant (Ka) to find x:
Ka = [C5H5NH+][OH-] / [C5H5N]
The value of Ka is given as pKa = -log(Ka) = 11.3, so we can write:
10^(-11.3) = (x)(x) / (0.08364 - x)
Solving this equation will give the concentration of pyridinium ion (C5H5NH+) at Va. Take the negative logarithm (pH) of this concentration to find the pH.
Step 3: Equivalence point (Ve)
The equivalence point occurs when the number of moles of HCl added is stoichiometrically equivalent to the number of moles of pyridine present initially. At this point, all of the pyridine has reacted to form pyridinium ions, and the solution becomes neutral.
Therefore, the pH at the equivalence point (Ve) should be approximately 7.
Step 4: Acid volume of 25.0 mL (after equivalence point)
At this point, we have added excess HCl past the equivalence point, meaning that we have an excess of H+ ions present. To calculate the pH, we need to determine the concentration of the excess H+ ions.
In this case, the excess H+ ions come from the dissociation of HCl. We know that the initial concentration of HCl is 0.1067 M, and we have added an additional 25.0 mL. Therefore, we need to calculate the concentration of H+, which is equal to the initial concentration of HCl minus the concentration of HCl that reacted with the pyridine.
Using an ICE table similar to the one in Step 2, we can determine the concentration of H+ ions and calculate the pH.
By following these steps and performing the necessary calculations, you should be able to obtain the correct pH values at each acid volume and compare them to the values you previously obtained.