Buffer solution preparation

[Nanu Nana]

Homework Statement

So I have prepare a buffer solution of 100 mL with ph of 4.00 .I have to use acetic acid and NaAc (sodium acetate ) to do so . The sum of the concentration of weak acid and its conjugate base is equal to 0.10 M
ca + cb = 0.10 M
How much volume of acetic acid and NaAc should I use to make buffersolution with ph of 4.00
Pka of acetic acid is = 4.76
Also information given :
NaAc (molar weight = 82.08g/mol ) and 100 % m
HAc (mw= 60.05) (density = 1.05 ) (100 % m)

Homework Equations

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ph = pka - log (HAc/Ac-)

The Attempt at a Solution

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I tried solving it but I'm not sure if its correct
4.00 =4.76 - log(HAc)/Ac-)
4.00-4.76 = - log ( Hac/ Ac-)
[HAc]/ [Ac-] = 10^0.76 = 5.754
we know ca +cb = 0.10 M x 0.1 L = 0.01 mol
nAc- + 5.754 x n Ac- = 0.01 mol
6.754 nAc- = 0.01 mol
so nAc- = 0.01/ 6.754 = 0.00148 mol nAc-
HAc = 0.01 - 0.00148 =0.008519 mol HAc
SO we have to use 0.01 mol of HAc and 0.001480 mol of NaAc
HAc
c=(100% x 10 x1.05 g/ml)/ 60.05 g/mol = 17.458 M
c= n/v
17.458 M= 0.01 mol/ V
V= 5.71 x10^-4 L => 0.57 ml We going to add 0.57 ml of HAc
and for NaAc = 0.001480 mol x Mw ( 82.08) = 0.121520 g NaAc
Is this correct ??Please explain if its not ??

 

[Borek]

HAc = 0.01 - 0.00148 =0.008519 mol HAc
SO we have to use 0.01 mol of HAc

0.008519 is not 0.01.

But in general you are one the right track (and the mass of sodium acetate looks OK to me).

 

[Nanu Nana]

0.008519 is not 0.01.

But in general you are one the right track (and the mass of sodium acetate looks OK to me).

But my teacher used 0.01 mol for HAc last year . Thats why i 'm confused

 

[Nanu Nana]

He used the total mol

 

[Borek]

Not sure what you mean.

You have correctly calculated you need 0.008519 moles of the acid, but in the next line 0.008519 miraculously became 0.01 moles. Why?

 

[Nanu Nana]

Not sure what you mean.

You have correctly calculated you need 0.008519 moles of the acid, but in the next line 0.008519 miraculously became 0.01 moles. Why?

The following buffer problem is solved by my teacher .
300 ml ph =9.90 c= ca+cb
We're going to use NaHCO3 and NaOH
ph = 9.90 = pka - log nHCO3-/ n CO32-
pka = 10.30
nHCo3-/ n co32- = 2.51
n HCO3- +n co32- = 0.500 x 300 ml = 150 mmol
2.51 n co32- + n co32- = 150
nco32- = 150/ 3.51 =42.7 mmol
How much NahCo and NAoh do we have to use to make that buffer solution
His answer was 150 mmol NaHCO3 and 42.7 mmol NaOH
Our teacher wrote this answer . Why did he use total mol ( 150 mmol) for NaHCO3 why not 150-42.7 mmol ??

 

[Borek]

These are different cases. In one you mix a conjugate acid and conjugate base, in the other you mix an acid with a base to neutralize it - and produce the conjugate base.

I guess your mistake is based on a common misunderstanding of the buffers. Can you tell what are the conjugate acid and base in each case?

 

[Nanu Nana]

Ac- is conjugate base and HAc weak acid
on second one
HC03- = acid
c032- = conjugate base

 

[Borek]

OK.

Where does the CO₃^2- (base) come from?

Where does the Ac^- (base) come from?

Do you see why these are different problems?

 

[Nanu Nana]

Yes Ac- came from Sodium acetate but on second case we added strong base NaoH . Naoh completely deprotonate weak acids .

 

[Borek]

So is it clear now why these questions are solved a bit differently, and why your solution is incorrect?

 

[Nanu Nana]

Yeah thank you very much