Which of the following could be added to a solution of acetic acid to prepare a
ID: 1002560 • Letter: W
Question
Which of the following could be added to a solution of acetic acid to prepare a buffer? sodium acetate sodium hydroxide nitric acid hydrofluoric acid sodium acetate only sodium acetate or sodium hydroxide hydrofluoric acid or nitric acid nitric acid only sodium hydroxide only A buffer that contains 0.110 M HY and 0.220 M Y has a pH of 8.77. What is the pH after 0.0010 mol of Ba(OH) is added to 0.750 L of solution? What species are in the buffer region of a weak acid-strong base titration? How are they different from the species at the equivalence point? Determine pH during the titration of 20.0 mL of 0.1000 M butanoic acid (Ka = 1.54 times 10) with 0.1000 M NaOH solutions after the following additions of titrant: 0 mL 10.00 mL 15.00 mL 19.95 mL 20.00 mL 20.50 mL 25.00 mL.Explanation / Answer
5 ) answer : b)
8)
millimoles of acid = 20 x 0.1 = 2
pKa = -log Ka
pKa = -log (1.54x 10^-5)
pKa = 4.81
a) 0.0 mL NaOH added
pH = 1/2 [pKa - log C]
pH = 1/2 [4.81 -log 0.1]
pH = 2.90
b) 10.00 mL added
it is half equivalence point here pH = pKa
pH = 4.81
c) 15.00 mL added
millimoles of base = 0.1 x 15 = 1.5
HA + NaOH --------------------> NaA + H2O
2 1.5 0 0 --------------> initial
0.5 0 1.5 - ---------------> after reaction
pH = pKa + log [NaA / HA]
pH = 4.81 + log (1.5 / 0.5)
pH = 5.29
d) 19.95 mL added
millimoles of base = 0.1 x 19.95 = 1.995
HA + NaOH --------------------> NaA + H2O
2 1.995 0 0 --------------> initial
0.005 0 1.995 - ---------------> after reaction
pH = pKa + log [NaA / HA]
pH = 4.81 + log (1.995 / 0.005)
pH = 7.41
e) 20.00 mL added :
it is equivalencepoint here only salt remains
salt molarity = 20 x 0.1 / (20 +20) = 0.05 M
pH = 7 + 1/2 [pKa +logC]
pH = 7 + 1/2 [4.81 + log 0.05]
pH = 8.75
f) 20.50 mL NaOH added
it is after the equivalence point so base remains
[OH-] = (20.50 x 0.1 - 20 x 0.1 ) / (20 +20.50)
= 1.23 x 10^-3 M
pOH = -log [OH-]
pOH = 2.91
pH + pOH = 14
pH = 11.09
g) 25 .00 mL NaOH added
it is after the equivalence point so base remains
[OH-] = (25 x 0.1 - 20 x 0.1 ) / (20 +25)
= 0.011 M
pOH = -log [OH-]
pOH = 1.95
pH + pOH = 14
pH = 12.05
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