Quiz V Ba(NO3)2 Ba2+ + 2NO NaCH3COO H2O Na+ + CH 3COO

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Quiz V
1) Label the following salts as neutral, acidic, or basic and write down the reactions that control
the acid-base behavior: Ba(NO3)2, NaCH3COO, NH4 Cl, K2SO4.
2O
Ba(NO 3 ) 2 "H"
# Ba 2+ + 2NO$3
From the above, we have the cation of a strong base, Ba(OH) 2, and the anion of a strong
acid, HNO3. Therefore, Ba(NO3)2 is neutral.
!
2O
NaCH 3COO "H"
# Na + + CH 3COO$
CH 3COO$ + H 2O % CH 3COOH + OH $
Na+ is the cation of a strong base, NaOH, and acetate is the anion of the weak acid acetic
acid. Thus, NaCH3 COO is a weak base.
!
2O
NH 4 Cl "H"
# NH +4 + Cl$
NH +4 + H 2O % NH 3 + H 3O +
Ammonium is the cation of a weak base, NH3, and chloride is the anion of a strong acid,
HCl. NH4Cl must be a weak acid.
!
2O
K 2SO 4 "H"
# 2K + + SO 4
2$
2$
$
SO 4 + H 2O % HSO 4 + OH $
!
Potassium is the cation of a strong base, sulfate is the anion of a weak acid , HSO4- .
K2SO4 is therefore a weak base. This next part was not necessary on the quiz, but to
convince you of this let’s look at it a bit closer. The Ka of the hydrogen sulfate ion is
1x10-2, so if one dissolved enough K2SO4 to make a 1.0 M solution the pH would be:
K w = K aK b
K b = K wK a = 10"14 M 2 /10"2 = 10"12 M
Kb =
[HSO"4 ][OH" ]
[SO 2"
4 ]
10"12 M =
x([OH" ]o + x) x(10"7 M + x)
x2
=
#
[SO 2"
1.0M " x
1.0M
4 ]o " x
x = [OH" ] = 3.2x10"6
[H + ][OH" ] = 10"14
[H + ] = 3.2x10"9
pH = 8.5, which is greater than 7 so the solution is basic
!
2) Consider two different acids, HA and HB, with the following acidities: pKa(HA) = 4 and
pKa(HB) = 5. What concentration ratios are needed to make a buffer with pH = 4.8? Which
of the two buffers has the higher capacity? Explain.
To do this problem, make use of the Hendersen-Hasselbach equation.
[base]
[acid]
[base]
pH " pK a = log
[acid]
[base]
= 10(pH"pK a )
[acid]
pH = pK a + log
For a pH of 4.8, and pK a of 4 and 5
[A]
= 10(4.8"4) = 6(.3)
[HA]
[B]
!
= 10(4.8"5) = 0.6(3)
[HB]
Where the parentheses indicate a non - significant digit
HB will have a higher capacity because the ratio above is closer to one, or equivalently
because its pKa is closest to the desired pH.
!
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