(b) (bh + )

11
(B) (BH + ) C 2 H 5 NH 2 (aq) + H 2 0 C 2 H 5 NH 3 + (aq) + OH - (aq) A weak base A 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 K b = [C 2 H 5 NH 3 + ] [OH - ] = (X) (X) = X 2 = 6.4 x 10 -4 [C 2 H 5 NH 2 ] (0.075-X) (0.075-X)

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(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 Presentation

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Page 1: (B)                                    (BH + )

(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)

Page 2: (B)                                    (BH + )

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

Page 3: (B)                                    (BH + )

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.

Page 4: (B)                                    (BH + )

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

Page 5: (B)                                    (BH + )

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

Page 6: (B)                                    (BH + )

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

Page 7: (B)                                    (BH + )

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]

Page 8: (B)                                    (BH + )

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

Page 9: (B)                                    (BH + )

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

Page 10: (B)                                    (BH + )

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

Page 11: (B)                                    (BH + )

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!