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.

 

[LudusRex]

It seems like you are on the right track with your calculations. However, there are a few things that could be causing your final answer to match the constant.

First, make sure you are using the correct value for the Kb of NH4+. You have listed it as 1.8*10^-5, but the correct value is actually 1.8*10^-5. This small change could make a big difference in your final answer.

Second, double check your units throughout your calculations. Make sure all units are consistent and that you are using the correct units for each value.

Third, when solving for x, make sure you are using the correct equation for an acid-base equilibrium. In this case, you are working with a weak base (NH3) and its conjugate acid (NH4+), so the equation should be Kb = [NH4+][OH-]/[NH3].

Lastly, check your final answer to make sure it makes sense in the context of the problem. In this case, the answer of 0.18M for NH4+ concentration seems reasonable given the initial concentration of NH3 and the desired pH of 9.

Overall, it's important to double check your work and make sure all values and calculations are accurate. Titrations and buffers can be tricky, so don't be afraid to ask for help or clarification if needed. Good luck on your test tomorrow!