A 1.00L buffer solution is formed by adding 0.250 moles of KOH (aq) to 0.300 moles of HA. The Ka 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?
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
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