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.

Answer 1

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