what is the pH in a 0.13M solution of aniline (Kb=4.3*10^-10).
What is the concentration of C6HNH2 in the same solution?
You made a typo in your post. Aniline is C6H5NH2
Let's call aniline a simple ANH2.
........ANH2 + HOH ==> ANH3^+ + OH^-
I.......0.13............0........0
C........-x.............x........x
E.......0.13-x..........x........x
Substitute the E line into the Kb expression for aniline and solve for x and 0.13-x. x = (OH^-), convert that to pOH then to pH.
To find the pH of a solution of aniline, we first need to determine the concentration of hydroxide ions (OH-) in the solution. Given that the Kb (base dissociation constant) of aniline is 4.3*10^-10, we can use the Kb expression to find the concentration of OH-.
Kb = [OH-][C6HNH2] / [C6HNH3+]
Since aniline is a weak base, we can assume that the concentration of OH- is equal to the concentration of C6HNH2 in the solution, so we can rewrite the equation as:
4.3*10^-10 = [C6HNH2]^2 / [C6HNH3+]
Now, let's find the concentration of C6HNH2 in the solution:
[C6HNH2]^2 = 4.3*10^-10 * [C6HNH3+]
[C6HNH2] = sqrt(4.3*10^-10 * [C6HNH3+])
Next, we need to find the concentration of C6HNH3+ in the solution. Given that the aniline solution is 0.13 M, we can assume complete ionization of the aniline, so the concentration of C6HNH3+ is also 0.13 M.
Now, substitute the value of [C6HNH3+] into the previous equation:
[C6HNH2] = sqrt(4.3*10^-10 * 0.13)
Calculate the concentration of C6HNH2 using a calculator:
[C6HNH2] = sqrt(5.59*10^-11)
[C6HNH2] ≈ 7.47 * 10^-6 M
Therefore, the concentration of C6HNH2 in the solution is approximately 7.47 * 10^-6 M.
Now, to find the pH of the solution, we need to know the concentration of H3O+ ions. Since the solution is slightly basic, we can assume that the concentration of H3O+ is the same as the concentration of OH-, which is 7.47 * 10^-6 M.
To find the pH, we can use the formula:
pH = -log[H3O+]
pH = -log(7.47 * 10^-6)
Calculate the pH using a calculator:
pH ≈ 5.13
Therefore, the pH of the 0.13 M solution of aniline is approximately 5.13.