(b) (bh + )
DESCRIPTION
(B) (BH + ) C 2 H 5 NH 2 (aq) + H 2 0 C 2 H 5 NH 3 + (aq) + OH - (aq). 16.75 by forum request. A conjugate acid of a weak base. A weak base. K b = [ C 2 H 5 NH 3 + ] [OH - ] = (X) (X) = X 2 = 6.4 x 10 -4 - PowerPoint PPT PresentationTRANSCRIPT
(B) (BH+)C2H5NH2(aq) + H20 C2H5NH3
+ (aq) + OH- (aq)
A weak baseA conjugate acid of
a weak base
16.75 by forum request
B BH+ OH-
0.075M 0 0
-X X X
(0.075-X) X X
Kb = [C2H5NH3+ ] [OH- ] = (X) (X) = X2 = 6.4 x 10-4
[C2H5NH2] (0.075-X) (0.075-X)
Kb = [C2H5NH3
+ ] [OH- ] = (X) (X) = X2 = 6.4 x 10-4
[C2H5NH2] (0.075-X) (0.075-X)
X2 = 6.4 x 10-4
(0.075-X)
X2 + 6.4 x 10-4 X – 4.8 x 10-5 = 0
X= -6.4x10-4 + (6.4x10-4)2 - 4(1) (-4.8 x 10-5) :X= 6.61 x 10-5 = [OH-] 2
6.61 x 10-5 = [OH-]
pOH- = 2.18, pH = 11.82
16.83 a NaCN Na+ + CN-
RECOGNIZE:THIS IS THE
CONJUGATE SALT OF A WEAK ACID
RECOGNIZE:THE CONJUGATE OF A STRONG BASE (GROUP
ONE METAL, NaOH) DOES NOT
HYDROLYZE, NO Ph EFFECT
RECOGNIZE:THE CONJUGATE OF A
WEAK ACID, (HAS A Ka) AND WILL
HYDROLYZE TO RAISE Ph RELEASING OH-
FROM HYDROLYSIS.
CN- + H2O OH- + HCN
HYDROLYSIS OF CN-, CONJUGATE BASE.
CONJUGATE BASE OF
HCN
WEAK ACID, IGNORE ANY IONIZATION AS IT IS INSIGNIFICENT.
CN- + H2O OH- + HCN
HYDROLYSIS OF CN-, CONJUGATE BASE.
CN- OH- HCN
0.10 0 0
-X X X
0.10-X X X
Kb = [HCN][OH-] WHERE Kb = Kw/Ka = 1 x 10-14/4.9x10-10= 2.0 x10-5
[CN-]
Look up the Ka of HCN
Kb = [HCN][OH-] = 2.0 x 10-5 = (X)(X) [CN-] (0.10-X)
2.0 x 10-5 = X2
(0.10)
FOR SAKE OF TIME
APPROXIMATE X
X2 = (0.10) 2.0x10-5
X = [OH-] =.00143pOH = 2.85; pH = 11.15
16.83 a
16.83 b: by forum request REACTION #1 : DISSOCIATION OF CONJUGATE SALT.
Na2CO3(aq) 2 Na+ (aq) + CO32-
(aq)
RECOGNISE THIS AS THE SALT OF A WEAK ACID AND STRONG (GROUP ONE METAL) BASE: NaOH
.REACTION #2 : HYDROLYSIS #1 OF CO3
2- , CONJUGATE BASE.
CO32- + H2O OH- + HCO3
-
THIS IS THE FIRST HYDROLYSIS OF THIS DIBASIC ION. Kb1 = Kw / Ka2
REACTION #3 : HYDROLYSIS #2 OF CO32- , CONJUGATE BASE.
HCO3- + H2O OH- + H2CO3
THIS IS THE SECOND HYDROLYSIS OF THIS DIBASIC ION. Kb2 = Kw / Ka1
Ka2<< Ka1 thereforeKb1>> Ka2, use Kb1 for pH
CO32- OH- HCO3
-
0.080 M 0 0
-X X X
0.080-X X X
AS USUAL, ASSESS THE STRONGEST IONIZATION FOR THE pH!
Kb = [HCO3-][OH-] WHERE Kb1 = Kw/Ka2 = 1 x 10-14/5.6 *10-11= 1.79 *10-4
[CO32-]
Kb = [HCO3-][OH-] = 1.79 * 10-4 = X2
[CO32-] 0.080-X
1.79 * 10-4 = X2
0.080-X
0.080 (1.79 * 10-4 ) = X2
X = 0.00378 = [OH-] :pOH- = 2.42 : pH = 11.58
16.83 b
REACTION #1 : DISSOCIATION Ka1 OF POLY (TRI) PROTIC ACID. ALL INOZATIONS ARE WEAK.
H3PO4(aq) H+ (aq) + H2PO4-
(aq)
RECOGNIZE THIS AS A WEAK POLYPROTIC ACID IONIZING IN PURE WATER.
.
16.113 MONSTER!
H3PO4 H+ H2PO4-
0.025 M 0 0
-X X X
0.025-X X X
Ka1 = [H+][H2PO4-]
[H3PO4]
7.5 X 10-3 = [X][X]
[0.025-X]
7.5 X 10-3 = [X][X]
[0.025-X]
7.5 X 10-3 = (X)2
(0.025-X)
0.025-X (7.5 X 10-3 ) = (X)2
X2 + (7.5 x 10-3X )– (1.875 x 10-4)=0
X= + 7.5 X 10-3 (7.5 X 10-3 )2 - 4(1) (- 1.875 x 10-4) : X= 0.01045 = [H+]
2(1) X= 0.01045 =[H2PO4- ]
16.113 MONSTER!
H2PO4- H+ HPO4
2-
0.01045 0.01045 0
-X X X
0.01045-X 0.01045 +X X
The H2PO4-
WILL DISSOCIATE IN
Ka2
Assume all H+ is from Ka1, it will appear in the ICE charts
of Ka2, Ka3
16.113 MONSTER!REACTION #2 : DISSOCIATION Ka2 OF POLY (TRI) PROTIC ACID. ALL INOZATIONS ARE WEAK.
H2PO4-(aq) H+ (aq) + HPO4
2- (aq)
RECOGNIZE THIS AS A WEAK POLYPROTIC ACID IONIZING IN PURE WATER.
.H2PO4- H+ HPO4
2-
0.01045 0.01045 0
-X X X 0.01045-X 0.01045 +X X
Ka2 = [H+][HPO42-]
[H2PO4-]
Ka2 = [0.010+X][X]
[0.010-X]
Ka2 = [0.010+X][X]
[0.010-X]
IGNORE ANY H+ PRODUCED BY Ka2,ASSUME ALL H+ IS
FROM Ka1
Ka2 = [0.010+X][X]
[0.010-X]
16.113 MONSTER!
Ka2 = 6.2 x 10-8 = 0.010X
0.010X = 6.2 X 10-8 = [HPO4
2-], THE SPECIES OF
INTEREST FROM Ka2 FOR Ka3
HPO42- H+ PO4
3-
6.2x10-8 0.01045 0
-X X X 6.2x10-8 -X 0.01045 +X X
HPO42- H+ PO4
3-
6.2x10-8 0.010 0
-X X X 6.2x10-8 -X 0.010 +X X
REACTION #3 : DISSOCIATION Ka3 OF POLY (TRI) PROTIC ACID. ALL INOZATIONS ARE WEAK.
HPO42-
(aq) H+ (aq) + PO43-
(aq)
RECOGNIZE THIS AS A WEAK POLYPROTIC ACID IONIZING IN PURE WATER.
.From Ka2
Ka2 = [H+][HPO42-]
[H2PO4-]
Ka2 = [0.010+X][X]
[6.2x10-8-X]
Ka3 = 4.2 x 10-13 = 0.010X
6.2x10-8
X = 2.6 x 10-18 = [PO43-], THE SPECIES OF
INTEREST FROM Ka3
16.113 MONSTER!