Determine the pH of a 1 M aniline solution (Kb = 4.2 10-10).
I am unsure how to start, any help?
Write the hydrolysis equation. Call aniline BNH2, then
........BNH2 + H2O ==> BNH3^+ + OH^-
I........1M.............0.......0
C........-x.............x.......x
E........1-x............x.......x
Kb = (x)(x)/(1-x)
Solve for x = OH and convert to pH.
To determine the pH of a 1 M aniline solution, you need to use the given value of Kb (base dissociation constant) for aniline.
Aniline (C6H5NH2) is a weak base, so it undergoes partial dissociation in water according to the following equation:
C6H5NH2 + H2O ⇌ C6H5NH3+ + OH-
The Kb expression for aniline is:
Kb = ([C6H5NH3+][OH-])/[C6H5NH2]
Given that the Kb value for aniline is 4.2 × 10^-10, we can assume that the concentration of OH- formed by the dissociation of aniline is very small compared to the original concentration of aniline.
To solve this problem, we'll make the approximation that [OH-] is negligible compared to the initial concentration of the aniline. This allows us to simplify the Kb expression to:
Kb = [C6H5NH3+]/[C6H5NH2]
Since aniline is a weak base, we can assume that the concentration of aniline before dissociation ([C6H5NH2]) is equal to the initial concentration of the solution (1 M). Thus, the equation becomes:
Kb = [C6H5NH3+]/1
Simplifying further gives:
[C6H5NH3+] = Kb
Now, we can use the given Kb value to find the concentration of C6H5NH3+:
[C6H5NH3+] = 4.2 × 10^-10
To determine the pH, we'll need to find the pOH first. The pOH is the negative logarithm (base 10) of the hydroxide ion concentration.
pOH = -log10([OH-])
Since we made the approximation that [OH-] is negligible compared to the initial concentration of aniline, we can assume that [OH-] ≈ 0. Hence, the pOH is approximately equal to 0.
Now, we can use the relationship between pH and pOH:
pH + pOH = 14
Since the pOH is approximately 0, we have:
pH + 0 = 14
Therefore, the pH of the 1 M aniline solution is approximately 14.