Question

Describe how to prepare 10 mL of 0.200 acetate buffer, pH 5.00, starting with pure acetic acid and solutions containg 3 M HCl and 3 M NaOH. Include both quantitate and qualitative details.

I know that a strong base reacts completely witha weak acid because the equilibrium constant in large. and that pH=pka + log ([NaOH]/[HCl])

Answer 1

acetate is the base. You need NaOH with HAc (acetic acid). You don't need the HCl.
pH = pKa + log (base)/(acid)
5.00 = pKa + log b/a. Look up pKa for HAc.
Solve for b/a ratio. That's eqn 1.
Equation 2 is b + a = 0.200

Solve those two equations simultaneously for b and a.
I prefer to work in millimols. You want 10 mL x 0.2M = 2 mmols.
Solving the equations give you base = 1.27 mmols and acid = 0.73 mmols.

.......HAc + OH^- ==> Ac^- + H2O
I.......2.....0........0.......
add..........1.27..............
C....-1.27..1.27.....1.27......
E......0.73..0.......1.27

So you want 0.73 mmols HAc and you want to add 1.27 mmols of 3M NaOH. M = mmols/mL or mL = mmols/M
mL NaOH = 1.27/3M = 0.423 mL

Pure HAc is 17 M. Solve the same way to see mL pure HAc. I have 0.118 mL
YOU SHOULD PLUG THOSE NUMBERS INTO THE HH EQUATION AND CONFIRM THAT IT GIVES YOU A Ph OF 5.00.
So you start with 0.118 mL of pure HAc, add 0.423 mL of 3M NaOH, make to slightly less than the mark of a 10 mL volumetric flask and check the pH. Add 3M NaOH or 3M HCl (small amounts) until the pH reads 5.00, then make to the mark of the volumetric flask. Mix thoroughly.

Answer 2

how do you prepare 100 mL of 0.200 M acetate buffer, pH5.00, starting with pure liquid acetic acid and solutions containing~3 M HCl and ~3 M NaOH

Answer 3

how to sagot that one huhu