Part A What is the pH of a buffer prepared by adding 0.607 mol of the weak acid
ID: 491198 • Letter: P
Question
Part A What is the pH of a buffer prepared by adding 0.607 mol of the weak acid HA to 0.507 mol of NaA in 2.00 L of solution? The dissociation constant K a of HA is 5.66× 10 7 . Express the pH numerically to three decimal places.
Part B What is the pH after 0.150 mol of HCl is added to the buffer from Part A? Assume no volume change on the addition of the acid. Express the pH numerically to three decimal places.
Part C What is the pH after 0.195 mol of NaOH is added to the buffer from Part A? Assume no volume change on the addition of the base. Express the pH numerically to three decimal places.
Explanation / Answer
We will need pKa for buffer:
pKa = -log(Ka) = -log(5.66*10^-7) = 6.2471
henderson haselbach equation
pH = pKa + log(A-/HA)
pH =6.2471 + log(0.507 /0.607 )
pH = 6.169
b)
pH after 0.15 mol of H+ is added
mol of HA = 0.607 + 0.150 = 0.757
mol of A- = 0.507 -0.150 = 0.357
substitute
pH = pKa + log(A-/HA)
pH =6.2471 + log(0.357/0.757)
pH = 5.921
c)
pH after 0.195 mol of OH- is added
mol of HA = 0.607 - 0.195 = 0.412
mol of A- = 0.507 +0.195 = 0.702
substitute
pH = pKa + log(A-/HA)
pH =6.2471 + log(0.702/0.412)
pH = 6.479