Question

Hello, I will try this again

I am trying to prepare a buffer from a solution of a weak acid.

This is for part D. of Experiment 25 - Ph Measurements - buffers and their properties

I am to be give an 0.50 M solution of weak acid with a given pKa, and I will preparre a buffer of a desired pH measurement (choose any pH)

I must first dilute the acid solution to 0.10 M by adding 10 ml of the acid to 40 ml of water.

Then calculate how much 0.10 M NaOH I must add to 20 ml of the acid.

I am having difficulty setting up my equation for this lab

D. Preparing a Buffer from a solution of a weak acid.

pH of buffer to be prepared: ___ (select pH), pKa of acid: ___ will be provided(select hypothetical)

Ratio of [B-]/[HB] required in buffer: ___

Volume of 0.10 M NaOH to be calculated ___ ml

pH of acid before titration:___

Initial volume reading of NaOH:___ml

Final volume reading of NaOH:__ml

Volume of NaOH actually required:

"Use the Henderson- hasselbalch formula to calculate the Ph of the buffer that is Ph = Pka + log{[salt]/[acid]"

I see the equation, can anyone work out the problem with my values??

Answer 1

I must first dilute the acid solution to 0.10 M by adding 10 ml of the acid to 40 ml of water.

Then calculate how much 0.10 M NaOH I must add to 20 ml of the acid.

since we have 0.1M NaOH

Acid is 0.1 M and 20 mL we will need 20 mL of 0.1M NaOH.

This will give full neutralization

On the other hand to prepare a buffer you will need a mixture of Acid and base

When Acid and base are 1:1 that means you have carried out 1/2 neutralization

In the earlier case after adding 10 mL of 0.1M NaOH to 20 mL 0.1M Acid

We will have [B-]/[HA] = 1

So from Henderson Hasselbalch equation

pH = pKa

If you want any other pH different from the pKa

If you want pH more than pKa then in the

Henderson Hasselbalch equation

pH = pKa + log base/acid

log base/acid has to be positive so base should be more than acid

whereas if pH shoule be less than pKa

base/acid should be a fraction so base should be less than acid

Since you have everything here in general terms hope you understand. Once you have pKa and pH it will be easier

pH = pKa + log base/acid

8.0 = 8.7 + log base/acid

log base/acid = -0.7

base/acid = 10^-0.7

base/acid = 0.199

base should be 0.199 of that of acid

0.1 M 20 mL acid then we have 0.1 x 0.02 = 0.002 mole

base + acid = 0.002

base = 0.002-acid

0.002-acid/acid = 0.199

0.002 - acid = 0.199acid

0.002 = 1.199 acid

acid = 0.002/1.199

acid = 0.001668

base = 0.002-0.001668 = 3.32 x 10^-4

since we have a 0.1 M NaOH we need to add 3.32 mL

to get the buffer of pH 8.0