A 25.0 mL sample of 0.100 M pyridine (Kb for pyridine is 1.7 ✕ 10-9) is titrated with 0.100 M HCl solution. Calculate the pH after the addition of the following amounts of HCl with sig figs. a) 24.5 mL b) 25.0 mL c) 26.0 mL d) 28.0 mL e) 30.0 mL
Oh, pH calculations! I hope you're ready for a titrating tale of acidity and basicity. Here we GO!
To start off, we have a solution of pyridine, but we're adding HCl to it. Pyridine is a weak base, so when we add HCl, it will react with the pyridine and cause the solution to become more acidic.
Now, let's break down each scenario and determine the pH after the addition of the specified amounts of HCl.
a) 24.5 mL: Since 24.5 mL is less than the initial volume of the pyridine solution, the solution remains basic. The pH will be greater than 7, so let's put on our clown noses and move on.
b) 25.0 mL: Ah, this is an interesting one. At this point, the volumes of pyridine and HCl are equal. We're at the equivalence point, folks! This means that all the pyridine has reacted with the HCl, resulting in a neutral solution. The pH will be perfectly 7, like a circus balancing act.
c) 26.0 mL: We've tipped the scales, my friend. The amount of HCl added has exceeded the volume of pyridine. The solution is now acidic. To calculate the pH, we need to find the concentration of HCl remaining in solution. Since the initial concentrations were the same, the concentration of HCl at this point is 0.100 M - (0.100 M/25 mL x 26 mL). Calculate that, and the pH will be a clownishly low number.
d) 28.0 mL: The acidity is ramping up, isn't it? By this point, we've significantly surpassed the initial pyridine volume. To find the remaining HCl concentration, we'll need to subtract the amount that reacted from the initial concentration. Do the math, and you'll find yourself with a pH that's getting closer to the depths of zero.
e) 30.0 mL: The grand finale! At this point, we've poured in even more HCl than before. The solution is now very acidic. Use the same method as before to calculate the pH, and brace yourself for a dive into acidic territory.
And that's it! Remember, pH calculations can be complicated, but with a little clownish humor and perseverance, you'll get through them. Happy titrating!
To calculate the pH at different points during the titration, we need to consider the reaction between pyridine (a weak base) and HCl (a strong acid).
Step 1: Calculate the moles of pyridine initially present.
Given:
Volume of pyridine solution (V1) = 25.0 mL
Concentration of pyridine (C1) = 0.100 M
Using the formula:
moles = concentration × volume (in liters)
moles of pyridine (n1) = C1 × V1
Substituting the values:
n1 = 0.100 M × 0.0250 L
n1 = 0.0025 mol
Step 2: Determine the limiting reagent and calculate the moles of excess HCl.
Given:
Concentration of HCl (C2) = 0.100 M
Volume of HCl added (V2) = volume of HCl - volume of pyridine (since pyridine is initially present)
a) V2 = 24.5 mL - 25.0 mL
b) V2 = 25.0 mL - 25.0 mL
c) V2 = 26.0 mL - 25.0 mL
d) V2 = 28.0 mL - 25.0 mL
e) V2 = 30.0 mL - 25.0 mL
The limiting reagent in this reaction is pyridine because HCl is in excess.
Step 3: Calculate the moles of pyridine remaining after the reaction.
moles of pyridine remaining (n2) = n1 - n(HCl)
The moles of HCl can be calculated using the formula:
moles = concentration × volume (in liters)
moles of HCl = C2 × V2
Substituting the values, we can now calculate the moles of pyridine remaining.
Step 4: Calculate the concentration of pyridine remaining.
Given:
Final volume of the solution (Vf) = V1 + V2
Using the formula:
concentration = moles / volume (in liters)
concentration of pyridine remaining (Cp) = n2 / Vf
Step 5: Calculate the pOH and then pH using the pOH.
Given:
Kb for pyridine = 1.7 × 10^-9
Using the formula:
Kw = Ka × Kb
Kw = 1.0 × 10^-14 (at 25°C)
pOH = -log10 (OH^- concentration)
OH^- concentration can be calculated using the formula:
OH^- concentration = sqrt(Kw / Cp)
pH = 14 - pOH
Now let's calculate the pH for each case:
a) V2 = 24.5 mL
b) V2 = 0 mL (equivalence point)
c) V2 = 26.0 mL
d) V2 = 28.0 mL
e) V2 = 30.0 mL
To calculate the pH after the addition of specific amounts of HCl, we need to determine the amount of pyridine remaining and the resulting concentration of its conjugate acid, pyridinium ion.
First, let's find the initial number of moles of pyridine:
moles of pyridine = volume of pyridine (L) × concentration of pyridine (M)
= (25.0 mL ÷ 1000 mL/L) × 0.100 M
= 0.00250 moles
Next, let's calculate the number of moles of HCl added:
moles of HCl = volume of HCl (L) × concentration of HCl (M)
a) For 24.5 mL:
moles of HCl = (24.5 mL ÷ 1000 mL/L) × 0.100 M
b) For 25.0 mL:
moles of HCl = (25.0 mL ÷ 1000 mL/L) × 0.100 M
c) For 26.0 mL:
moles of HCl = (26.0 mL ÷ 1000 mL/L) × 0.100 M
d) For 28.0 mL:
moles of HCl = (28.0 mL ÷ 1000 mL/L) × 0.100 M
e) For 30.0 mL:
moles of HCl = (30.0 mL ÷ 1000 mL/L) × 0.100 M
Now, let's determine the remaining moles of pyridine after each addition of HCl:
remaining moles of pyridine = initial moles of pyridine - moles of HCl added
To calculate the concentration of pyridine (M) after each addition, divide the remaining moles of pyridine by the final volume (in liters) after the HCl addition.
For each scenario (a-e), the concentration of pyridine remaining post-HCl addition is given by:
concentration of pyridine = remaining moles of pyridine ÷ (final volume of solution in mL ÷ 1000 mL/L)
Finally, we can calculate the concentration of the pyridinium ion (conjugate acid):
concentration of pyridinium ion = initial concentration of pyridine - remaining concentration of pyridine
To find the pH, we need to use the equation for the Kb of pyridine:
Kb = [pyridinium ion][OH-] / [pyridine]
We can assume that the concentration of hydroxide ion ([OH-]) is equal to the concentration of hydronium ion ([H+]) since the solution is acidic. Therefore, we have:
[H+] = Kb × [pyridine] / [pyridinium ion]
pH = -log10[H+]
By plugging in the values of [H+] we found above into the pH equation, we can calculate the pH after the addition of each amount of HCl.
You have five problems here. You must recognize where you are on the titration curve to each of these. Pyridine is C5H5N.
.............C5H5N + HCl ==> C5H5N*HCl
millimols C5H5N initially = mL x M = 25.0 x 0.1 = 2.50
a. millimols HCl added. 24.5 x 0.1 = 2.45
b. millimols HCl added. 25.0 x 0.1 = 2.50
c. millimols HCl added. 26.0 x 0.1 = 2.60
d. millimols HCl added. 28.0 x 0.1 = 2.80
e. millimols HCl added. 30.0 x 0.1 = 3.00
The equivalence point is at b. So a is before the eq. pt. and c,d,e are after.
a. Before the eq. pt. you will still have some pyridine not neutralized and you will have formed some of the salt. You have a buffered solution.Use the Henderson-Hasselbalch equation. pH = pKa + log (base)/(acid)
.............C5H5N + HCl ==> C5H5N*HCl
I................25.0.......0..............0
add......................24.5.......................
C.............-24.5...-24.5.........24.5
E..............0.5...........0..........24.5
pH = pKa + log (base)/(acid)
pH = pKa + log (0.5/24.5)
b. The equivalence point. The pH is determined by the hydrolysis of the salt which is as follows: The total volume now is 25.0 + 25.0 = 50.0 so the concn of the C5H5N*HCl to start is M = mmols/mL = 2.50/50 = 0.05 M
.................C5H5N*HCl + HOH ==> C5H5N + H3O^+ + Cl^-
I..................0.05.................................0..............0.............0
C..................-x....................................x..............x
E.............0.05-x...................................x..............x
Ka for C5H5N*HCl =Kw/Kb for C6H6N) = (x)(x)/(0.05-x) = (1E-14/1.79E-9)
Solve for x = (H3O^+) and convert that to pH.
c,d,e. All are AFTER the eq. pt so you will have neutralized ALL of the pyridine and you will have an excess of the acid . I will do the c and leave d and e to you.
.............C5H5N + HCl ==> C5H5N*HCl
I..............2.50..........0..............0.
add.........................2.60...........................
C..........-2.50........-2.50.............2.50
E.............0.............0.1.............2.50
The M of the excess HCl. You have 0.1 mmols HCl and you have a volume of 25.0 + 26.0 = 51 mL so M = 0.1/51 = ?. Since that is straight HCl, that gives you the (H^+) and convert that to pH.
d and e are done the same way with the only difference being the total volume which for 28 mL added titrant is 25.0 + 28 and the other is 25.0 + 30.0.
Post your work if you get stuck on any of these. A blackboard is useful for showing this because you can see the titration curve and see where you are with each of these additions.
a.OOOPS. I substituted volumes and not mmols for the a part. I should have done this. Before the eq. pt. you will still have some pyridine not neutralized and you will have formed some of the salt. You have a buffered solution.Use the Henderson-Hasselbalch equation. pH = pKa + log (base)/(acid)
.............C5H5N + HCl ==> C5H5N*HCl
I................2.50.......0.............0..............0
add......................2.45.......................
C.............-2.45...-2.45.........2.45
E..............0.05...........0..........2.45
pH = pKa + log (base)/(acid)
pH = pKa + log (0.05/2.45)
I don't think that changes the answer any since both numerator and denominator were too high by a factor of 10 so that cancels and the answer stays the same; however, the actual numbers were wrong and I didn't like that. Sorry about the mixup