[Boxxxed]

I am calculating how many ml of 0.1 M CH3COOH is needed to make a buffer with a pH of 5.80

ph=5.80=4.75+log(base/acid)

11.22=base/acid

11.22(acid)=base

11.22(acid)=0.1M-acid

How did they go from the above step to this?

12.22(acid)=0.1M

They gave the acid a value of one and added it to both sides, Why?

[Borek]

I am calculating how many ml of 0.1 M CH3COOH is needed to make a buffer with a pH of 5.80

Is it a complete question? You need both acid and conjugate base, don't you?

[Boxxxed]

Yes, I forgot to add. It asks for ml of both conjugate base and acid.

0.1 M of the conjugate base is given.

[Boxxxed]

I forgot to mention we are trying to create a 100 ml buffer.

5.80=4.75(base/acid)

11.22=base/acid

Total Moles of buffer = (0.1M)(0.1L) = 0.01 moles

0.01moles = 11.22(acid)+(acid)
0.01moles = 12.22(acid)
acid = 8.18x10^-4 moles

The answer is 8.18x10^-3

Figured it out.

8.18x10^-4 moles divided by 0.1L = 8.18x10^-3