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Introduction to Ka, Kw and Kb (Acids and Bases)
Goes through a shorthand method of determining Ka, Kw and Kb of chemicals
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Kw Ka and KbUsing ,
Kw =
Ka -
Kb Kb
=
¥0T , where KW =
1.0 ✗ 10
-
"t
this for acid-base pair
is
only true
conjugate
1×7 What is the Kb following WA 's ?
for the
-0×10
14/5.6×10-10
'
-
Nffg
=
1.8×10-5 =
conj .
acid Ntlyt
" "°
" "
/ 4.5×10-7
"
two -3
= =
2.2×10-8
conj acid 1-12103
.
""
0%4.7×10-11
"
coz =
=
2.1×10-4
acid HCG
-
conj .
lx) what is the pH of 0.10
Moye HCl
? 0 10m01 / l
.
CHSCOOH ?
Hzot -10g 10.10) 1.00
-
=
HCl t
tho → ta
1-130++4-1300
-
Cttscootl tho
lH3¥¥YJ
+
⇒ Ka
-
-
A Ct
0.10 Moye
-
'S
-
x + ✗ TX 1.8×10 =
approx VUU 0.10 -
✗
0 10m01
.
-
✗ ✗ ✗
0.00134M 011L 1.8×10-6 =
✗
2
1094-130+3
-
0.00134 MOI / L =
X
pH
=
2.87
lx) Of HF is dissolved in 1.00L of water
50.0g (g) . What
sound
'
pH of this solution?
is the
so.ogi-o.mg
'
=
F- H2O in ⇒
tbitaq,
HFIQ, +
Fiaq)
250m01/ L 0 -0
6.3×10-4=2 ,go¥- approx
-
2.sound l
/ X ✗ ✗
-
PH 0-03968 moth :X
-10g(Hzgt)
:
1- 401
=
SA :
ionize 1001 , ,
no ka needed initial conc
=
11-130+1
WA :
use ICE tables ! Ka initial cone > 11-1307 (only small percentage 151-6017
SB :
dissociate 100% , no Kb needed initial cone [01-1]*1include mole ratio)
=
-
-
0.10 NaOH →
Nat + OH ←
0.10
"
Mg(OH)a
→ -
0.10
Mg t 201-1 ← 0.20
-
WB
:
use Ky and ICE tables initial conc > 101-1-1
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