# Homework Help: Figuring pH of .20M NH3 and .35M NH4Cl

1. Oct 31, 2012

### mesa

1. The problem statement, all variables and given/known data
Calculate the pH of a solution that is .20M NH3 and .35M NH4Cl with a Kb(NH3)=1.8x10^-5

3. The attempt at a solution
So first calculated the M of OH- produced by the NH3,
[OH-]=√((1.8x10^(-5))x.20) = 1.9x10^-3

Since NH4Cl will dissociate in H2O it will leave NH4+, NH4+ + H2O ---> NH3 + H30+
Ka would be [1x10^(-14)]/[1.8x10^(-5)] = 5.6x10^(-10) giving us a rather insignificant amount of hydronium produced compared to the amount of OH- from the NH3 but lets figure it none the less,

[H3O] = √(5.6x10^(-10))x.35 = 1.4x10^(-5)
Since the hydronium will react with the OH- subtracting the two gives us 1.88x10^-3M OH-
giving us a pOH of 2.73, or pH of 11.27

The answer is pH 9.01 so I am waaaaay off.
This is new stuff for my brain and that's the excuse I'm sticking with :)
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Oct 31, 2012

### Staff: Mentor

This is a buffer solution.

3. Oct 31, 2012

### mesa

I was breaking it down a little further than the Henderson–Hasselbalch equation so I can get a better conceptual understanding.

The following is correct for that equation is it not?

pH = pKa + log (base/acid)
so pH = 4.74 + log [1.9x10^(-3)]/[5.6x10^-10]

I still get 11.27pH, yikes!

I must be screwing up on the molarity, do you see where I put my marbles on this thing Borek?

4. Oct 31, 2012

### Staff: Mentor

Your problem is with the identification of the base and its conjugate acid. These are not OH- and H+.

5. Oct 31, 2012

### mesa

Blamo, Marbles found! :)