This question I really don't get it. I tried tried tried to figure out how to start solve the problem but still couldn't figure out. So can you help me with one question PLEASE? If so, can you please explain to me how you get the answer? Also can you show me step by step of solving the problem? I need to know how to do this problem, I will have test on Monday so I need to understand it. Thank!

Here the question:

what is the pH of a solution that is 0.30 M in aniniline, C6H5NH2 and 0.15M in anilinium chloride, C6H5NH3Cl? Show the equation for ionziation

Have you had a class on the Henderson-Hasselbalch equation?

pH = pKa + log[(base)/(acid)]
The base is aniline (0.30M) and the acid is the chloride(0.15M). The pKa is the pKa of aniline. In my book the Kb for aniline is 3.94E-10 which makes pKb = 9.40 and that subtracted from 14 makes pKa = 4.60.
The ionization equation for aniline is
C6H5NH2 + HOH ==> C6H5NH3^+ + OH^-

No, my professor didn't teach student on Henderson-Hasselbalch equation. My professor only show us the Henderson-Hasselblach forumla but not show how to use it. Also she didn't give us what is pKa for this problem. Anyway, thank you so much for your help.

How did you get the 3.94E-10?

As I stated in my response I looked up Kb for aniline in one of my texts and obtained 3.94E-10. You should look it up in your text; your text may have a slightly different value. However, -log Kb = pKb; therefore, pKb for aniline from my value is 9.40 and since pKa + pKb = 14, then pKa = 14-9.40 = 4.60.

Then substitute into the HH equation this way,
pH = 4.60 + log(0.30)/(0.15) and solve for pH. I get 4.90.

There is another way to do it and not use the HH equation.
....C6H5NH2 + HOH ==> C6H5NH3^+ + OH^-
I.....0.30.............0...........0
C......-x...............x..........x
E.....0.30-x.............x.........x

Then Kb = (C6H5NH3^+)(OH^-)/(6H5NH2)

....C6H5NH3Cl^- ==> C6H5NH3^+ + Cl^-
I....0.15..........0..........0
C...-0.15..........0.15.......0.15
E......0..........0.15.........0.15

Now substitute into the Kb equation this way:
(C6H5NH3^+) = x + 0.15 (x from the ionization of C6H5NH2 and 0.15 from the anilinium chloride.
(OH^-) = x
(C6H5NH2) = (0.30-x) and it look this way.

3.94E-10 = (x+0.15)(x)/(0.30-x)
Assume x+0.15 = 0.15 and assume 0.30-x = 0.30. You can do that since x is small. That gives you
3.94E-10 = (0.15)(x)/(0.30) and solve for x which is OH^-.
I get 7.88E-10 for OH and pOH = 9.10 and pH = 14-9.10 = 4.90.
You can see why we use the Henderson-Hasselbalch equation; it is so much easier to use for buffer problems.

Yea, I realized that HH is more easy and faster. I'm going use HH better instead other way. Anyway thank you so much for explain.

Of course, I'll be glad to help you with your question!

To find the pH of the solution containing aniline and anilinium chloride, we need to consider the ionization equation for aniline. Aniline, C6H5NH2, acts as a weak base in water and ionizes to form its conjugate acid, anilinium ion (C6H5NH3+), and hydroxide ion (OH-).

The ionization equation for aniline can be written as follows:

C6H5NH2 + H2O ⇌ C6H5NH3+ + OH-

Now, let's consider the concentrations given in the problem. The solution is 0.30 M in aniline (C6H5NH2) and 0.15 M in anilinium chloride (C6H5NH3Cl). The anilinium chloride dissociates in water to form anilinium ion and chloride ion (Cl-).

To find the pH of the solution, we need to calculate the concentration of hydroxide ions (OH-) in the solution. Since we have the concentration of anilinium chloride, we can assume that the concentration of anilinium ion (C6H5NH3+) is equal to 0.15 M.

Now, we can use the ionization equation to determine the concentration of hydroxide ions. Since aniline is a weak base, we can assume that the concentration of hydroxide ions is equal to the concentration of anilinium ions.

Therefore, [OH-] = 0.15 M.

Next, to find the pH, we need to calculate the concentration of H+ ions (or H3O+ ions). We can use the equation:

Kw = [H+][OH-]

where Kw is the ion product constant for water, which is 1.0 × 10^-14 at 25°C.

Given that [OH-] = 0.15 M, we can rearrange the equation to solve for [H+]:

[H+] = Kw / [OH-]

[H+] = (1.0 × 10^-14) / (0.15)

Calculating this value will give us the concentration of H+ ions.

Finally, to find the pH, we can use the equation:

pH = -log[H+]

Substituting the calculated value of [H+], we can determine the pH of the solution.

I hope this explanation helps you understand how to approach and solve the problem step by step. If you have any additional questions, feel free to ask!