Question

Your boss wants you to make some buffers using the acetic acid / acetate buffering system. You find 0.100 M stock solutions of Na acetate and acetic acid on the shelf. Use your resources to find the pKa for this system and calculate hte amount in mL of each you would have to use to make 500 mL of the following buffer

Desirec pH : 4.00

mL Na acetate _____

mL acetic acid _____

please show work

Answer 1

Solution:

pka of acetic acid = 4.75

Given : Total volume of the buffer = 500 mL

pH of the buffer = 4.0

Molarity of stock solution = 0.100 M

We know [Na acetate]= [ acetate ion]

We use Henderson-Hasselbalch equation to get ratio of [acetate]/[acetic acid]

The equation is

pH = pka + log ([acetate ion]/[acetic acid])

Lets plug the values

4.00 = 4.75 + log ([acetate ion]/[acetic acid])

-0.75 = log ([acetate ion]/[acetic acid])

Lets take antilog of both side

0.178 = ([acetate ion]/[acetic acid]

Volume is same so we can use following equation

([acetate ion]/[acetic acid])

= moles of acetate ion/moles of acetic acid

moles of acetate ion/moles of acetic acid = 0.178

Lets assume moles of acetate ions = x and that of acetic acid is y

x / y = 0.178

x = 0.178 y

Now we also know

Volume of acetate ion + volume of acetic acid = 500 mL

Volume = mol / molarity

Lets plug this equation

0.178y / 0.100 + y / 0.100 = 0.500 L

Lets plug value of x

0.178 y + y = 0.500 L* 0.100

1.178 y = 0.05

y = 0.04244

y = moles of acetic acid

x = 0.178 * y = 0.178 * 0.04244 = 0.0075 mol

Now we know the equation

Volume of acetate ions = x / 0.1 M

= 0.0075 mol / 0.1 M = 0.0755 L

= 75.5 mL

Volume of acetic acid = 500-75.5 mL = 424.5 mL