Pyridine is a weak base that is used in the manufacture of pesticides and plastic resins. It is also a component of cigarette smoke. Pyridine ionizes in water as follows:
C5H5N + H2O<-->C5H5NH+ + OH-
The pKb of pyridine is 8.75. What is the pH of a 0.190 M solution of pyridine?
Well, looks like pyridine likes to hang out in water and have a little ionizing party. But before we dive into the pH, let's take a step back and understand what's going on here.
Pyridine is a weak base, which means it only partially ionizes in water. In this case, it steals a proton (H+) from water to become C5H5NH+, also known as pyridinium, and leaves behind an OH- ion. It's like a game of musical chairs, but with chemical compounds instead.
Now, the pKb of pyridine is 8.75. To find the pH of a solution, we need to convert pKb into pKa, which is the negative logarithm of the acid dissociation constant. Don't worry, I'll do the math for you. Just sit back and enjoy the ride!
pKb + pKa = 14
Given that pKb = 8.75, we can solve for pKa:
pKa = 14 - 8.75
pKa = 5.25
Now that we have the pKa value, we can calculate the concentration of pyridinium ions ([C5H5NH+]) in the solution:
[C5H5NH+] = √((Ka * [C5H5N]) / [OH-])
We know that, for water at 25°C, [H2O] = 1 × 10^-7 M, so [OH-] = 1 × 10^-7 M.
Plugging in the values, we get:
[C5H5NH+] = √((10^-14 * 0.190) / (1 × 10^-7))
[C5H5NH+] = √((10^-14 * 0.190) * 10^7)
[C5H5NH+] ≈ √(1.9 × 10^-6)
[C5H5NH+] ≈ 0.00138 M
Now, let's move on to the pH calculation. Since we have the concentration of pyridinium ions, we can assume the concentration of pyridine is approximately the same (we're dealing with a weak base here, remember?). So the concentration of pyridine ([C5H5N]) is also 0.00138 M.
pH = -log[H+]
But wait, we don't have [H+] directly. No worries! We can assume that the [H+] concentration is equal to the [OH-] concentration in this case. So:
[H+] = 1 × 10^-7 M
Plugging in the values, we get:
pH = -log(1 × 10^-7)
pH ≈ 7
Voila! The pH of a 0.190 M solution of pyridine is approximately 7. So, if you were planning on having a pyridine party, make sure to bring a neutral pH punch!
To find the pH of a 0.190 M solution of pyridine, we need to determine the concentration of hydroxide ions (OH-) in the solution first. Then we can use the concentration of hydroxide ions to calculate the pOH and finally convert it to pH.
Given that the pKb of pyridine is 8.75, we can calculate the Kb (base dissociation constant) using the equation:
Kb = 10^(-pKb)
Kb = 10^(-8.75)
Kb = 1.78 x 10^(-9)
Since the pyridine reacts with water to form hydroxide ions (OH-), the concentration of OH- can be calculated using the Kb value and the initial concentration of pyridine.
Let's assume "x" represents the concentration of OH- ions formed.
C5H5N + H2O<-->C5H5NH+ + OH-
Initially, 0.190 M pyridine ionizes to form x M of hydroxide ions and x M of pyridinium ions.
At equilibrium, the concentration of pyridine is (0.190 - x) M, and the concentration of hydroxide ions is x M.
The Kb expression for pyridine can be written as:
Kb = [C5H5NH+][OH-]/[C5H5N]
Using the equilibrium concentrations, the equation can be written as:
1.78 x 10^(-9) = x * (0.190 - x) / (0.190 - x)
Simplifying the expression:
1.78 x 10^(-9) = x^2 / (0.190 - x)
Since x is small compared to 0.190, we can approximate (0.190 - x) as 0.190 in the denominator.
1.78 x 10^(-9) = x^2 / 0.190
Solving for x:
x^2 = 1.78 x 10^(-9) * 0.190
x^2 = 3.382 x 10^(-10)
Taking the square root of both sides:
x = sqrt(3.382 x 10^(-10))
x ≈ 1.84 x 10^(-5)
Now, we can calculate the pOH of the solution using the concentration of hydroxide ions:
pOH = -log10[x]
pOH = -log10[1.84 x 10^(-5)]
pOH ≈ -(-4.733)
pOH ≈ 4.733
Finally, we can calculate the pH using pOH:
pH = 14 - pOH
pH = 14 - 4.733
pH ≈ 9.267
Therefore, the pH of a 0.190 M solution of pyridine is approximately 9.267.
To find the pH of a 0.190 M solution of pyridine, we need to determine the concentration of hydroxide ions (OH-) resulting from the ionization of pyridine in water. This can be done by using the pKb value of pyridine.
The pKb is defined as the negative logarithm of the base dissociation constant (Kb). In this case, the pKb of pyridine is 8.75. Recall that the pKb is related to the Kb by the equation:
pKb = -log(Kb)
To find Kb, we can rearrange the equation:
Kb = 10^(-pKb)
Plugging in the value of pKb for pyridine:
Kb = 10^(-8.75)
Now, let's calculate the concentration of hydroxide ions using the Kb value and the initial concentration of pyridine (0.190 M):
Since pyridine is a weak base, we can assume that at equilibrium, most of the pyridine will remain unchanged, and only a small fraction will react with water to form hydroxide ions.
Let's assume that x M of pyridine ionizes:
C5H5N(aq) + H2O(l) ⇌ C5H5NH+(aq) + OH-(aq)
The initial concentration of C5H5N is 0.190 M, and after ionization, the concentration of OH- is also x M.
From the balanced equation, we can see that the concentration of OH- is equal to x M.
Using the Ka expression for the reaction, we have:
Kb = [C5H5NH+][OH-] / [C5H5N]
Plugging in the known values:
10^(-8.75) = x * x / (0.190 - x)
Since x is small compared to 0.190, we can assume that (0.190 - x) ≈ 0.190. This allows us to simplify the equation:
10^(-8.75) = x^2 / 0.190
Rearranging the equation, we have:
x^2 = (10^(-8.75)) * 0.190
Taking the square root of both sides:
x ≈ √[(10^(-8.75)) * 0.190]
Calculating this expression will give us the approximate value of x, which represents the concentration of hydroxide ions (OH-). Finally, we can use this concentration to find the pOH and convert it to the pH:
pOH = -log[OH-]
pH = 14 - pOH
pKb = -log Kb
-8.75 = log Kb
Solve for Kb.
........BN + H2O ==> BNH^+ + OH^-
I......0.190..........0.......0
C.......-x............x........x
E......0.190-x........x........x
Substitute the E line into the Kb expression and solve for x = OH^-, then convert to pH.