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
I answered a similar problem early today but it three more point on the titration curve. Your two points are the same. Follow up with the post.
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To calculate the pH after the addition of a certain amount of HCl, we need to determine the concentration of HCl and the concentration of the conjugate acid of pyridine, pyridinium ion.
First, let's calculate the moles of pyridine initially present in the 25.0 mL sample:
moles of pyridine = volume (L) x concentration (mol/L)
moles of pyridine = (25.0 mL / 1000 mL/L) x 0.100 mol/L
moles of pyridine = 0.00250 mol
a) After adding 24.5 mL of 0.100 M HCl:
The moles of HCl added = volume (L) x concentration (mol/L)
moles of HCl added = (24.5 mL / 1000 mL/L) x 0.100 mol/L
moles of HCl added = 0.00245 mol
The reaction between pyridine and HCl is:
C5H5N (pyridine) + HCl ⇌ C5H6N+ (pyridinium ion) + Cl-
Since the ratio of reactants is 1:1, the moles of pyridine that react with HCl is also 0.00245 mol.
Now, we need to calculate the moles of pyridine left:
moles of pyridine left = initial moles of pyridine - moles of pyridine reacted with HCl
moles of pyridine left = 0.00250 mol - 0.00245 mol
moles of pyridine left = 5.0 × 10^(-5) mol
To calculate the concentration of pyridinium ion:
volume of pyridinium ion = volume of HCl added (24.5 mL) + initial volume of pyridine (25.0 mL)
volume of pyridinium ion = (24.5 mL + 25.0 mL) / 1000 mL/L
volume of pyridinium ion = 0.0495 L
concentration of pyridinium ion = moles of pyridinium ion / volume of pyridinium ion
concentration of pyridinium ion = 0.00245 mol / 0.0495 L
concentration of pyridinium ion = 0.0495 M
To calculate the pOH of the solution:
pOH = -log10[OH-]
pOH = -log10(Kb x [pyridinium ion])
pOH = -log10(1.7 x 10^(-9) x 0.0495)
pOH ≈ 7.44
Next, we can calculate the pH using the pOH:
pH = 14 - pOH
pH = 14 - 7.44
pH ≈ 6.56
Therefore, after the addition of 24.5 mL of 0.100 M HCl, the pH of the solution is approximately 6.56 (with sig figs).
b) After adding 25.0 mL of 0.100 M HCl:
Since the volume of HCl added is equal to the initial volume of pyridine, no pyridine is left in the solution. This means the solution consists only of the conjugate acid of pyridine, pyridinium ion.
To calculate the concentration of pyridinium ion:
concentration of pyridinium ion = moles of pyridinium ion / volume of HCl added
concentration of pyridinium ion = 0.00250 mol / 0.0250 L
concentration of pyridinium ion = 0.100 M
To calculate the pOH of the solution:
pOH = -log10[OH-]
pOH = -log10(Kb x [pyridinium ion])
pOH = -log10(1.7 x 10^(-9) x 0.100)
pOH ≈ 6.43
Next, we can calculate the pH using the pOH:
pH = 14 - pOH
pH = 14 - 6.43
pH ≈ 7.57
Therefore, after the addition of 25.0 mL of 0.100 M HCl, the pH of the solution is approximately 7.57 (with sig figs).
To calculate the pH after the addition of the indicated amounts of HCl, we need to consider the reaction between pyridine (C5H5N) and HCl. Pyridine is a weak base, and when it reacts with HCl, it forms the pyridinium cation (C5H5NH+) and chloride ion (Cl-).
The balanced chemical equation for the reaction is as follows:
C5H5N + HCl -> C5H5NH+ + Cl-
To determine the pH, we need to examine the resulting solution and determine the concentration of each species. Here's how you can do it for each case:
a) After adding 24.5 mL of 0.100 M HCl:
1. Calculate the number of moles of HCl added:
moles HCl = (volume of HCl in liters) x (molarity of HCl)
= (24.5 mL / 1000 mL/L) x (0.100 mol/L)
= 2.45 x 10^-3 mol HCl
2. Use stoichiometry to determine the number of moles of pyridine that reacted with HCl. Since the reaction is 1:1,
moles pyridine reacted = 2.45 x 10^-3 mol HCl
3. Subtract the moles of pyridine reacted from the initial moles of pyridine to find the remaining moles of pyridine.
moles pyridine remaining = initial moles of pyridine - moles pyridine reacted
4. Calculate the remaining concentration of pyridine by dividing the remaining moles by the final volume in liters (25.0 mL + 24.5 mL).
concentration pyridine = (moles pyridine remaining) / (total volume in liters)
= (initial moles of pyridine - moles pyridine reacted) / (49.0 mL / 1000 mL/L)
= (0.0250 mol - 2.45 x 10^-3 mol) / (0.0490 L)
= [0.025 - 0.00245] / 0.0490
= 0.0131 M
5. Calculate the pOH using the equation pOH = -log10[OH-]. Since pyridine is a weak base and Kb for pyridine (C5H5N) is given as 1.7 x 10^-9, we can assume that all the pyridine reacts with HCl to give an equal concentration of [C5H5NH+] and [OH-].
Kb = [C5H5NH+][OH-]/[C5H5N]
1.7 x 10^-9 = (x)(x) / (0.0131)
x^2 = (1.7 x 10^-9) x (0.0131)
x = sqrt((1.7 x 10^-9) x (0.0131))
x ≈ 3.27 x 10^-6 M
6. Use the pOH to calculate the pH using the equation pH = 14 - pOH:
pH = 14 - (-log10(3.27 x 10^-6))
≈ 8.49
Therefore, the pH after adding 24.5 mL of HCl is approximately 8.49.
b) After adding 25.0 mL of 0.100 M HCl:
Following the same steps as above, you'll find that the concentration of pyridine remaining is zero (0.0131 mol - 0.00131 mol = 0). Since there is no pyridine left, the calculation for the pH is not valid.
Therefore, after adding 25.0 mL of HCl, the pH cannot be determined as there is no remaining pyridine to affect the solution.