# PH calculation

1. May 24, 2005

### Guillermo

Here is the problem:

Given pKa=9.3 what is pH of 0.005 mol/l ammonia solution.

The problem is, I don't get this pKa part. I mean, I know t means Ka=5.0119*10^-10 but ammonia is a base, not acid.

G

2. May 24, 2005

### Staff: Mentor

pKa + pKb = pKw

Last edited by a moderator: Aug 13, 2013
3. May 24, 2005

### Guillermo

I am a moron :( We spent more than hour discussing it yesterday.

OK, so Kb=1.9953*10^-5.

But there is still something wrong.Ammonia is a weak base, so pH=(0.005*Kb)^0.5=3.5? This is way too low.

G

4. May 24, 2005

### Staff: Mentor

No. What you have calculated is pOH, not pH. Besides, check the degree of dissociation - are you sure you can use simplified equation?

Last edited by a moderator: Aug 13, 2013
5. May 24, 2005

### Guillermo

Ah, so the pH is 10.5!

G

6. May 24, 2005

### Staff: Mentor

NO!

Equation you have used - [OH] = sqrt(C*Kb) is valid only if the degree of dissociation is below 5%. In this case it is higher:

3.16e-4(dissociated ammonia concentration)/0.005(ammonia concentration) * 100% = 6.3%

so you have to use full quadratic equation.

Smarkotan oz gluthozmaz

Last edited by a moderator: Aug 13, 2013
7. May 27, 2005

### Guillermo

I missed your last post and my result was wrong

G

PS Check my new problem, please...

8. May 27, 2005

### GCT

alright, take it step by step...to go from pKa to Ka, simply negative inverse log of pKa. KaKb=Kw, Kb=Kw/Ka (Kw is the autodissociation of water).

$$Kb=[OH-][NH4+]/[NH3],~Kb=[x][x]/[initial~conc.~NH3-x]$$ solve for x, and that will be [0H-]. pOH=-log[0H-], pH+p0H=14, solve for pH.

that's it, you shouldn't be having so much trouble with this