.IU TlO CH3NH3 a) 1,111, and V b) III andv c) II, IV, and V d) IV and V e) Vonly
ID: 1033768 • Letter: #
Question
.IU TlO CH3NH3 a) 1,111, and V b) III andv c) II, IV, and V d) IV and V e) Vonly omeWook Boses 20 hoose appropriate acid/base pair for a buffer with a specific pH 31) Which acid/base conjugate pair will be best to make a buffer with a pH of 5? a) b) c) d) e) HCN/NaCN (Ka for HCN = 4.9 x 10 10) CsH5NHCl/C5H5N (Kb for CsHsN = 1.8 x 10")? HNO2/NaNO2 (Ka for HNO2 4.5 x 104) N2HsBr/N2H4 (Kb for N2Ha = 8.9 x 10") HOI/KOI (Ka for HOI = 2.3 x 10-11) 32) Which acid would be the best choice for a buffer with a pH 6? a) Formic acid K, - 1.8 x 10437 b) Hydrazoic Ka - 1.9 x 1051.9 c) Ascorbic Ka-8.0 x 105 y d) Hydrosulfuric Ka 1.0x 10 7 e) Arsenious K, 6 x 10101 09 8? 33) Which of the following bases will be best to use to create a buffer with pH ) NH3 Kb 1.8 x 10 L b) CeHsNH2 Kb = 4.3 x 10-10 c) H2NNH2 Kb 1.3 x 1o- d) C2HsNH2 Kb 6.4 x 104 e) NH2OH Kb = 9.1 x 109Explanation / Answer
A buffer pH should be closer to pKa of acid in case of acidic buffer
A buffer pH should be closer to pKa of conjugate acid in case of basic buffer
31)
pH = 5
so,
pKa should be closer to 5
use:
pKa = -log Ka
5.0 = -log Ka
Ka = 1*10^-5
use:
Kb = Kw/Ka
Kw is dissociation constant of water whose value is 1.0*10^-14 at 25 oC
Kb = (1.0*10^-14)/Ka
Kb = (1.0*10^-14)/1*10^-5
Kb = 1*10^-9
So,
Ka should be closer to 1*10^-5 or Kb should be closer to 1*10^-9
This is true for b
Answer: b
32)
pH = 6
so,
pKa should be closer to 6
use:
pKa = -log Ka
6.0 = -log Ka
Ka = 1*10^-6
Answer should be option c
33)
pH = 8
so,
pKa should be closer to 8
use:
pKa = -log Ka
8.0 = -log Ka
Ka = 1*10^-8
use:
Kb = Kw/Ka
Kw is dissociation constant of water whose value is 1.0*10^-14 at 25 oC
Kb = (1.0*10^-14)/Ka
Kb = (1.0*10^-14)/1*10^-8
Kb = 1*10^-6
Answer: c
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