Question

Using 1.8*10^-5 for the Ka for acetic acid, along with the weight of sodium acetate and volume of acetic acid used to prepare a buffer, calculate the expected pH of the buffer.

So the weight is 4.0089 g and the volume is 100mL

Answer 1

m = 4.0089 g of NaAc

MW of NaAc = 82.0343 g/mol

mol of NaAc = mass/MW = 4.0089 / 82.0343 = 0.04887 mol of NaAc

V =100 ml or 0.1 L

Concnetration of NaAc = mol/V = 0.04887 mol / 0.1 = 0.4887 M of NaAc

Note that in solution

NaAc ---> Na+ and Ac-

Therefore, 1 mol of NaAc will give 1 mol of Ac-

This will form a buffer

A buffer is:

HAc <---> H+ and Ac-

NaAc ---> Na+ and Ac-

you can see that Ac- is a common ion, will shift the equilibrium

Buffers are best described with Henderson Hasselbach equation

pH = pKa + log(A-/HA)

NOTE that we do not have a concnetration of HA or mass:volume ratio

I will assume HA = 0.1 which is pretty typical

for pKa

pKa = -log(Ka) = -log(1.8*10^-5) = 4.75

therefore:

pH = pKa + log(A-/HA) = 4.75 + log(0.4887/ 0.1) = 5.44

pH = 5.44

NOTE that I used a concnetration of HA = 0.1; please add your concnetration to the formula and oyu shoudl get the pH for that given concnetration