Question

calculate the ph when 1.59 g of ch3coona (fw = 82.03 g/mol) is added to 32.6 ml of 0.500 m acetic acid, ch3cooh. ignore any changes in volume. the ka value for ch3cooh is 1.8 x 10^-5.

Answer 1

Answer:4.81

Step-by-step explanation:

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 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

CREDIT GOES TO Ph.D. in Biochemistry--University Professor--Chemistry Tutor J.R. S.