Using 1.8x10-5 for the Ka for acetic acid, along with the weight of sodium acetate and volume of acetic acid used to prepare the buffer, calculate the expected pH of the buffer. Show your work for full credit. How well does it compare to the observed value

pH = 4.95

Explanation:

The buffer was created using 4.0082g sodium acetate, mixed with 10.0 mL of 3.0 M acetic acid and 90 mL of deionized water.

Using H-H equation for acetic buffer:

pH = pKa + log [Acetate] / [Acetic acid]

Where pH is the pH of the buffer

pKa = -log Ka = 4.74

[] Are molar concentration of each specie -The volume of the solution is 100mL = 0.1L-:

[Sodium acetate] = 4.0082g * (1mol / 82.03g) = 0.04886 moles / 0.1L

= 0.4886M

[Acetic acid] = 3.0M * (10mL / 100mL) = 0.300M

pH = 4.74 + log [0.4886M] / [0.300M] =

first one is the correct answer