What is the pH of a solution that is 0.50 M in sodium acetate and 0.75 M in acetic acid?

Question

Cameron Andrews · Accepted Answer

4,57

this is an acid tampon solution. #pH = pKa + log ((Cs)/(Ca))# = 4,75 + log(0,5/0,75) =4,75 +log 0,66 = 4,57 # pKa is the logarithm (changed of sign) of the dissociation constant of acetic acid; Cs the concentration of the salt (sodium acetate) ; Ca the concentration of the acid (acetic acid). As you can see pH doesn't change with concetration, and change very little changing the amount of salt and of acid

Ellie Andrews · Answer

undefined To determine the pH of the solution, we can use the Henderson-Hasselbalch equation:

pH = pKa + log([A-]/[HA])

Given that sodium acetate is the conjugate base of acetic acid, and the Ka of acetic acid is approximately 1.8 x 10^-5, we can calculate the pKa:

pKa = -log(Ka) = -log(1.8 x 10^-5) ≈ 4.74

Now, we can plug in the values into the Henderson-Hasselbalch equation:

pH = 4.74 + log(0.50/0.75) ≈ 4.74 + log(2/3) ≈ 4.74 - 0.176 ≈ 4.56

Therefore, the pH of the solution is approximately 4.56.