[aestas]

Calculate the pH of a mixture of 50 cm3 of ammonia solution of concentration
0.10 mol dm−3 and 50 cm3 of hydrochloric acid solution of concentration
0.050 mol dm−3 .

We're only given the pKb of ammonia at 4.75.

[enahs]

Can you calculate the pH of the two seperate solutions?


What is the balanced equation for the reaction? What is left over and how much? What is the pH of that solution?

[aestas]

Alright this is what I have...

for ammonia

Kb = [NH4+] [OH-]/[NH3]

[OH-]= square root {(10^-4.75)(0.1)}

[OH-]= 0.001333521 mol

1.0x 10^14 = [OH-][H+]

[H+] = 7.499 x 10^12 mol

for HCl

C=n/v

n = 0.05x0.05
n = 0.0025 mol

to find pH
0.0025mol + 7.499 x 10^12 mol = 0.0025? -> that makes the pH ~ 3

The answer is supposed to be 9.25...

[enahs]

My first question was just asking if you could calculate the pH if the solutions where seperate, just as a primer.


They are not seperate.
The next part with the balanced equation was the real hint. And the point where I was what is left over and how much.

Acids and bases neutralize each other to form water. You are going to form water and use either all the acid or all the base (not true if they are both weak though).

Which one do you have more of? Which one will be used up? What will you have left and what is the pH of that solution?

[aestas]

Would it be possible to use the pKb in the Henderson-Hasselbach formula in place of the pKa?

The formula is NH3 + HCl makes NH4+ + Cl-...

Ahh I see... you form ammonium and no water. So the HCl doesn't really play a factor in this and the pH is actually just the pH of the ammonium.

I wrote out the equation while working, but didn't see the light until I was just typing this post. I guess I just answered my first question. God, I feel so dumb for wasting so much time trying to figure out a question that was so obvious.

Thanks for the help.

[Borek]

Would it be possible to use the pKb in the Henderson-Hasselbach formula in place of the pKa?

This is exactly what you are supposed to do

[AWK]

And a final result will be pH=14-pKb=9,25 without HH equation. Why?