What is the pH of a 0.41 M solution of pyridinium chloride (C5H5NH+Cl−)? (Kb for pyridine is 1.5×10−9.)
Answer in units of pH
Ignore J's response. Just plug into DrBob222's explanation
.....C5H5NH^+ + HOH ==> C5H5N + H3O^+
I.....0.41M...............0.......0
C......-x.................x.......x
E.....0.41-x..............x.......x
Ka for C4H5NH^+ = Kw/Kb for pyradine = (C5H5N)(H3O^+)/(C5H5NH^+)
Plug in Kw/Kb to find Ka, then solve for x = (H3O^+). Then use pH = -log(H3O^+) for find (H3O^+).
Post your work if you get stuck.
To find the pH of a solution of pyridinium chloride (C5H5NH+Cl−), we need to determine the concentration of hydroxide ions (OH-) in the solution first.
Pyridine (C5H5N) is a weak base, and we have its Kb value, which is 1.5×10−9. Kb represents the base dissociation constant, and it tells us how much of the base dissociates in water.
The Kb expression for pyridine can be written as follows:
Kb = [OH-][C5H5NH+]/[C5H5N]
In this case, we have the concentration of pyridinium chloride (0.41 M), which is equal to the concentration of pyridinium ions ([C5H5NH+]). We'll assume that the concentration of pyridine ([C5H5N]) is negligible compared to this value.
Since we're trying to find [OH-], we can rearrange the equation as follows:
Kb = [OH-][C5H5NH+]/[C5H5N]
[OH-] = (Kb * [C5H5N])/[C5H5NH+]
Now we can substitute the given values into the equation:
[OH-] = (1.5×10−9 * 0.41 M)/0.41 M
[OH-] = 1.5×10−9
To find the pH, we need to convert the concentration of hydroxide ions ([OH-]) into the concentration of hydrogen ions ([H+]). We can use the fact that in water, the concentration of [H+] multiplied by [OH-] always equals 1 x 10^-14 (the ion product constant of water).
[H+] * [OH-] = 1 x 10^-14
[H+] = (1 x 10^-14)/[OH-]
[H+] = (1 x 10^-14)/(1.5×10−9)
[H+] = 6.67 x 10^-6 M
Now we can calculate the pH using the formula:
pH = -log10([H+])
pH = -log10(6.67 x 10^-6)
pH ≈ 5.18
Therefore, the pH of a 0.41 M solution of pyridinium chloride (C5H5NH+Cl−) is approximately 5.18.