How Do You Calculate the Required Amounts for Buffer Solutions?

[Taryn]

Hey I have been having a lot of trouble in solving these problems and I have exams soon and I really don't know where to start, I know u aint meant to give the answers so instead would someone please go through all of the steps to finish this!
I really need some kinda help, This question was in a quiz and I was so confused as to wat I needa do... I am going to get help from a lecturer as well but inorder to study it I need some help please.

A buffer solution of pH = 5.30 can be prepared by dissolving acetic acid and sodium acetate in water. How many moles of sodium acetate must be added to 1.0 L of 0.25 M acetic acid to prepare the buffer?
The answer is meant to be 0.90mol but I have no Idea how to get to that!

Also one more We mix 100 mL of 0.20 M HBr and 50.0 mL of 0.40 M NaC1O. What is the pH of the resulting solution? Ka(HC1O) = 3.5 x 10-8

for this one i Wrote out the equation... and found that the concentration of HClO was 0.1333 or something like that... then I just had no idea where to go from there, I no that pH=pka- log(acid/base) but I just didnt no how to get to that point... I changed Ka(HClO) to pKa by using -log3.5 x 10- 8 but as I sed not sure where to go.

Could anyone please help

 

[Hootenanny]

A buffer solution of pH = 5.30 can be prepared by dissolving acetic acid and sodium acetate in water. How many moles of sodium acetate must be added to 1.0 L of 0.25 M acetic acid to prepare the buffer?
The answer is meant to be 0.90mol but I have no Idea how to get to that!

Do you know how to calculate the acidity consant for buffer solutions?

 

[Taryn]

do you mean Kw and Kb coz I think so

 

[Hootenanny]

do you mean Kw and Kb coz I think so

I mean K_a perhaps you use different notantion? Anyway, the relationship between K_a and [H⁺ is given by;

K_{a} = [H^{+}] \times \frac{[\text{salt}]}{[\text{acid}]}

Where K_a is the acidity constant of the acid.

 

[Taryn]

I will give it a go... but i wat has me a little stumped is what happens to the pH, I obviously have to use that somewhere don't I?

 

[Taryn]

wait a second I understand now I think, I will work it all out and if I have a problem I will just post it on this agen, thanks a lot.

 

[Hootenanny]

I will give it a go... but i wat has me a little stumped is what happens to the pH, I obviously have to use that somewhere don't I?

You'll need that to work out the hydrogen ion concentration, you'll also need to look up K_a for acetic acid.

 

[Taryn]

okay sorry just one more question, I am a little confused... just coz wat will be the [H+], I wrote this, 1.8E-5=[H+] x [salt]/[o.25]

Then after I find the [salt] then I will just use n=c* volume right?
If you can't give me anymore help that's fine!

 

[Hootenanny]

okay sorry just one more question, I am a little confused... just coz wat will be the [H+], I wrote this, 1.8E-5=[H+] x [salt]/[o.25]

That looks good to me. You need to use you pH value to calculate the [H⁺], remember that;

pH = -log[H^{+}] \Leftrightarrow [H^{+}] = 10^{-pH}

Then after I find the [salt] then I will just use n=c* volume right?
If you can't give me anymore help that's fine!

Yes, you would usually do that, but since the solution is dissolve in one litre, the calculation becomes n = c x 1 and therefore in this case; n = c

Do you follow?

 

[Borek]

This equation (after taking log of both sides) is called Henderson-Hasselbalch equation. Check also buffer example 1 and buffer example 2.

 

[Taryn]

thanks very much that was heaps helpful and now I understand wat to do, don't know y I didnt think of that at the time.