Titrations and buffers help, test tommorow

[Nelo]

I don't understand how to do this problem, but i Understand how to do the reverse of it.

"You have 25mL of 0.10M nh3 solution. How many grams of nh4Cl do you need to add to make a buffer with a ph of 9?) answer is 0.18M of nh4+.

What I've done is set up the initial table.
nh3 + h20 ----> nh4+ oh-
I 0.10 0.10
C-x +x x

edit::(this table is posting wrong.. but you get the point... x is on nh4 and oh h2o is neglected)
E0.10-10^-9 (keeping values same basically)

Second thing i did was solve x by doign the antilog of the -ph.
giving me 1*10^-9. which i believe is correct.

kb for nh4 is 1.8*10^-5.

I set up this : 1.8*10^-5 = 0.10x/ 0.10

End up getting the same answer as the constant. Cant be right, what am i doing wrong when the ph of the buffer is given?

 

[SpectraCat]

I don't understand how to do this problem, but i Understand how to do the reverse of it.

"You have 25mL of 0.10M nh3 solution. How many grams of nh4Cl do you need to add to make a buffer with a ph of 9?) answer is 0.18M of nh4+.

What I've done is set up the initial table.
nh3 + h20 ----> nh4+ oh-
I 0.10 0.10
C-x +x x

edit::(this table is posting wrong.. but you get the point... x is on nh4 and oh h2o is neglected)
E0.10-10^-9 (keeping values same basically)

Second thing i did was solve x by doign the antilog of the -ph.
giving me 1*10^-9. which i believe is correct.

kb for nh4 is 1.8*10^-5.

I set up this : 1.8*10^-5 = 0.10x/ 0.10

End up getting the same answer as the constant. Cant be right, what am i doing wrong when the ph of the buffer is given?

Have you covered the Henderson Hasselbach equation in class?

 

[Nelo]

No, we have not. we don't use pKa values at all. We simply have covered simple buffers. However I don't understand how to solve this question.. and apparently normal buffer questions when mol/L are different

 

[Borek]

Using ICE tables for buffer calculations - while sometimes possible - is a waste of time. There is a much better tool for that - Henderson-Hasselbalch equation.

This question is quite easy to solve - convert Kb to pKa (pKa + pKb = 14, pKb = -log(Kb)), plug everything given into HH equation, and solve for the only unknown - [HN₄⁺].

Please don't ignore capital letters in formulas, they are there to avoid ambiguity. co doesn't meant anything as CO and Co are completely different things.