Consider a solution that contains both C5H5N and C5H5NHNO3. (Kb for pyridine, C5H5N, is 4.0 x 10^-4)
(a) Calculate the ratio of [C5H5N]/[C5H5NH+] if the solution has a pH of 4.50.
(b) Calculate the ratio of [C5H5N]/[C5H5NH+] if the solution has a pH of 5.00.
Oh, you want to talk about chemistry? Well, let me put on my lab coat and goggles. Let's dive into it!
(a) Alright, to calculate the ratio of [C5H5N]/[C5H5NH+], we'll need to use the pH of the solution and the pKb of pyridine (C5H5N). But before we get into math, let me tell you a hilarious pyridine joke. What did the pyridine say to the acidic solution? "Would you please stop pH-ing so mean to me?" Haha, classic!
Now, back to business. To find the ratio, we'll use the Henderson-Hasselbalch equation, which is pH = pKa + log [base]/[acid]. Since Kb = 4.0 x 10^-4, we can find pKb by taking the negative log of Kb.
pKb = -log (4.0 x 10^-4) = 3.4
Now, let's plug everything into the equation. Since pH = 4.50 and pKb = 3.4, we have:
4.50 = 3.4 + log [C5H5N]/[C5H5NH+]
Rearranging the equation, we get:
log [C5H5N]/[C5H5NH+] = 4.50 - 3.4 = 1.10
To get rid of the logarithm, we need to raise 10 to the power of both sides. So:
[C5H5N]/[C5H5NH+] = 10^1.10
Calculating that, we find:
[C5H5N]/[C5H5NH+] ≈ 12.6
So, the ratio of [C5H5N]/[C5H5NH+] in the solution with a pH of 4.50 is approximately 12.6.
(b) Now, for the solution with a pH of 5.00. We'll use the same equation and pKb value, but a different pH.
5.00 = 3.4 + log [C5H5N]/[C5H5NH+]
Rearranging the equation:
log [C5H5N]/[C5H5NH+] = 5.00 - 3.4 = 1.60
Taking 10 to the power of both sides:
[C5H5N]/[C5H5NH+] = 10^1.60
Calculating that, we find:
[C5H5N]/[C5H5NH+] ≈ 39.8
So, the ratio of [C5H5N]/[C5H5NH+] in the solution with a pH of 5.00 is approximately 39.8.
And there you have it! I hope my funny antics didn't distract you too much. If you have any more chemistry questions, feel free to ask!
To solve these problems, we will use the Henderson-Hasselbalch equation:
pH = pKa + log([A-]/[HA])
In this equation, [A-] represents the concentration of the conjugate base and [HA] represents the concentration of the weak acid.
For pyridine (C5H5N), it acts as a weak base, and its conjugate acid is C5H5NH+. The pKa value for this acid-base pair is determined by the Kb value:
pKa = 14 - pKb
(a) pH = 4.50
pKa = 14 - pKb = 14 - (-log(Kb)) = 14 - (-log(4.0 x 10^-4))
pKa = 14 - (-3.60) = 17.60
Using the Henderson-Hasselbalch equation:
4.50 = 17.60 + log([C5H5N]/[C5H5NH+])
Subtracting 17.60 from both sides:
-13.10 = log([C5H5N]/[C5H5NH+])
Converting to exponential form:
10^(-13.10) = [C5H5N]/[C5H5NH+]
Therefore, the ratio of [C5H5N]/[C5H5NH+] when the solution has a pH of 4.50 is approximately 7.943 x 10^(-14).
(b) pH = 5.00
pKa = 14 - pKb = 14 - (-log(Kb)) = 14 - (-log(4.0 x 10^-4))
pKa = 14 - (-3.60) = 17.60
Using the Henderson-Hasselbalch equation:
5.00 = 17.60 + log([C5H5N]/[C5H5NH+])
Subtracting 17.60 from both sides:
-12.60 = log([C5H5N]/[C5H5NH+])
Converting to exponential form:
10^(-12.60) = [C5H5N]/[C5H5NH+]
Therefore, the ratio of [C5H5N]/[C5H5NH+] when the solution has a pH of 5.00 is approximately 3.981 x 10^(-13).
To solve this problem, you need to use the Henderson-Hasselbalch equation, which relates the pH of a solution to the ratios of the concentrations of a weak acid and its conjugate base. The equation is as follows:
pH = pKa + log([base]/[acid])
In this case, pyridine (C5H5N) is the weak base, and pyridinium ion (C5H5NH+) is its conjugate acid. The pKa value can be calculated using the pKw and Kb values:
pKw = pKa + pKb
Given that the Kb for pyridine is 4.0 x 10^-4, we can calculate the pKb value as follows:
pKb = -log(Kb)
pKb = -log(4.0 x 10^-4)
Now we can calculate the pKa value using the pKw equation:
pKw = 14
pKa + pKb = 14
Substituting the calculated pKb value into the equation:
pKa + (-log(4.0 x 10^-4)) = 14
Now solve for pKa:
pKa = 14 + log(4.0 x 10^-4)
Now that we have the pKa value, we can proceed to calculate the ratios [C5H5N]/[C5H5NH+] for the given pH values:
(a) If the solution has a pH of 4.50, substitute the variables into the Henderson-Hasselbalch equation:
4.50 = pKa + log([C5H5N]/[C5H5NH+])
Solving for log([C5H5N]/[C5H5NH+]):
4.50 - pKa = log([C5H5N]/[C5H5NH+])
Substituting the pKa value we calculated earlier:
4.50 - (14 + log(4.0 x 10^-4)) = log([C5H5N]/[C5H5NH+])
Now solve for [C5H5N]/[C5H5NH+] by taking the antilog:
[C5H5N]/[C5H5NH+] = antilog(4.50 - (14 + log(4.0 x 10^-4)))
(b) If the solution has a pH of 5.00, repeat the steps above with the new pH value:
5.00 = pKa + log([C5H5N]/[C5H5NH+])
Solving for log([C5H5N]/[C5H5NH+]):
5.00 - pKa = log([C5H5N]/[C5H5NH+])
Substituting the calculated pKa value:
5.00 - (14 + log(4.0 x 10^-4)) = log([C5H5N]/[C5H5NH+])
Solve for [C5H5N]/[C5H5NH+] by taking the antilog:
[C5H5N]/[C5H5NH+] = antilog(5.00 - (14 + log(4.0 x 10^-4)))
By following these steps, you should be able to solve the problem and obtain the ratios [C5H5N]/[C5H5NH+] for the given pH values.
C5H5N + HOH ==> C5H5NH^+ + OH^-
Ka = (Kw/Kb) = (C5H5NH^+)(OH^-)/(C5H5N).
You know Kw and Kb. From pH you can calculate (OH^-). That leaves the ratio of C5H5N/C5H5NH^+ to calculate. The reciprocal is what you are looking for. Part b is done the same way.