Calculate the pH when 1.61 g of CH3COONa (FW = 82.03 g/mol) is added to 34 mL of 0.500 M acetic acid, CH3COOH. Ignore any changes in volume. The Ka value for CH3COOH is 1.8 x 10-5.

Question

Calculate the pH when 1.61 g of CH₃COONa (FW = 82.03 g/mol) is added to 34 mL of 0.500 M acetic acid, CH₃COOH. Ignore any changes in volume. The K_a value for CH₃COOH is 1.8 x 10^-5.

The answer is 4.81

please explain your steps, thank you :)

Answer 1

Again, as in your previous question, this is a buffer problem. You have a buffer because you have a weak acid (CH3COOH) and the salt of that acid (CH3COONa). We can again use the Henderson Hasselbalch equation for a weak acid buffer which is pH = pKa + log [salt]/[acid]

Before plugging in values, let us first calculate the pKa for acetic acid:

pKa = -log Ka = -log 1.8x10^-5 = 4.745

Next, we will need to calculate the concentration of the salt, i.e. [CH3COONa]:

1.61 g x 1 mol/82.03 g = 0.01963 moles in 34 ml (0.034 L) = 0.01963 moles/0.034L = 0.5773 M

Finally, we can now plug these values into the HH equation and find the pH:

pH = 4.745 + log [0.5773]/[0.500]

pH = 4.745 + log 1.155

pH = 4.745 + 0.0624

pH = 4.807 = 4.81