Question

A 1.00L buffer solution is formed by adding 0.250 moles of KOH (aq) to 0.300 moles of HA. The K_a for HA = 3.1 x 10^-5.

^{A. What is the initial pH of the buffer?}

^{B. What will pH be after the addition of 0.0500 moles of H3O+?}

^{C. 10ml of a 3.5 M solution of NaOH are added to the buffer. What is the resulting pH?}

Answer 1

a)

initial pH for buffer

pKa = -log(Ka) = -log(3.1*10^-5) = 4.51

pH = pKa + log(A-/HA)

mol of conjugate formed = 0.25

mol of acid left = 0.30-0.25 = 0.05

pH = 4.51+ log(0.25/0.05) = 5.2089

B)

after ading m

moles of H+ = 0.05

mol of conjugate reacted = 0.25 -0.05 = 0.20

mol of acid formed = 0.05+0.05 = 0.10

pH = 4.51+ log(0.20/0.10) = 4.81102

c)

if we add

mol of NaOH = MV = (10*10^-3)(3.5) = 0.035 mol of OH-

note; assume this was added to the latest buffer

if we add

initially

mol of conjugate 0.20

mol of acid = 0.10

after addition of base

mol of conjugate 0.20 + 0.035 = 0.235

mol of acid = 0.10 -0.035 = 0.065

substitute

pH = 4.51 + log(0.235/0.065) = 5.0681