Question

Prepare 100.0 mL of a 1.00 M pH 5.00 buffer using solid sodium acetate (FW = 82.031 g mol–1 ) and 6.00 M aqueous acetic acid (Ka = 1.78 x 10–5 ). How many mL of acetic acid and how many g of sodium acetate are needed?

So far I have:

pka=4.75 5= 4.75 +log [A]/[HA]

.25= log [A]/[HA]

1.78=[A]/[HA]

Answer 1

pH = pka + log [A-]/[hA}

5 = 4.75 + log [A-]/[HA]

[A-] = 1.78 [HA] ..............(1)

we have buffer concentration 1M , hence [A-] +[HA] = 1M .............(2)

Using (1) in ( 2) we get 1.78 [HA] + [HA] = 1 ,

[HA] = 0.36 ,   hence [A-] = 0.64 M

buffer volume is 100 ml = 0.1 L

Hence moles of HA = M x V = 0.36 x 0.1 = 0.036

Moles of A_ needed = 0.64 x 0.1 = 0.064

Sodium acetate needed in grams = Moles x molar mass = 0.064 x 82.031 = 5.25 g

acetic acid concentration = moles / vol in L

6 = 0.036 /vol

vol of acetic acid = 0.006 L = 6 ml