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A conjugate acid of a weak base
16.75 by forum request (B) (BH+) C2H5NH2(aq) + H20 C2H5NH3+ (aq) + OH- (aq) A conjugate acid of a weak base A weak base B BH+ OH- 0.075M -X X (0.075-X) Kb = [C2H5NH3+ ] [OH- ] = (X) (X) = X = 6.4 x 10-4 [C2H5NH2] (0.075-X) (0.075-X)
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Kb = [C2H5NH3+ ] [OH- ] = (X) (X) = X2 = 6.4 x 10-4
[C2H5NH2] (0.075-X) (0.075-X) X2 = x 10-4 (0.075-X) X x 10-4 X – 4.8 x 10-5 = 0 X= -6.4x (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
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NaCN Na+ + CN- CN- + H2O OH- + HCN RECOGNIZE: 16.83 a RECOGNIZE:
THE CONJUGATE OF A WEAK ACID, (HAS A Ka) AND WILL HYDROLYZE TO RAISE Ph RELEASING OH- FROM HYDROLYSIS. 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 CN- + H2O OH- + HCN HYDROLYSIS OF CN-, CONJUGATE BASE. WEAK ACID, IGNORE ANY IONIZATION AS IT IS INSIGNIFICENT. CONJUGATE BASE OF HCN
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HYDROLYSIS OF CN-, CONJUGATE BASE. FOR SAKE OF TIME APPROXIMATE X
CN- + H2O OH- + HCN HYDROLYSIS OF CN-, CONJUGATE BASE. 16.83 a CN- OH- HCN 0.10 -X X 0.10-X Look up the Ka of HCN Kb = [HCN][OH-] WHERE Kb = Kw/Ka = 1 x 10-14/4.9x10-10= 2.0 x10-5 [CN-] FOR SAKE OF TIME APPROXIMATE X Kb = [HCN][OH-] = 2.0 x 10-5 = (X)(X) [CN-] (0.10-X) 2.0 x 10-5 = X2 (0.10) X2 = (0.10) 2.0x10-5 X = [OH-] =.00143 pOH = 2.85; pH = 11.15
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Na2CO3(aq) 2 Na+ (aq) + CO32- (aq)
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 CO32- , CONJUGATE BASE. CO32- + H2O OH- + HCO3- THIS IS THE FIRST HYDROLYSIS OF THIS DIBASIC ION. Kb1 = Kw / Ka2 Ka2<< Ka1 therefore Kb1>> Ka2, use Kb1 for pH REACTION #3 : HYDROLYSIS #2 OF CO32- , CONJUGATE BASE. HCO3- + H2O OH- + H2CO3 THIS IS THE SECOND HYDROLYSIS OF THIS DIBASIC ION. Kb2 = Kw / Ka1
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16.83 b AS USUAL, ASSESS THE STRONGEST IONIZATION FOR THE pH! CO32- OH- HCO3- 0.080 M -X X 0.080-X Kb = [HCO3-][OH-] WHERE Kb1 = Kw/Ka2 = 1 x 10-14/5.6 *10-11= 1.79 *10-4 [CO32-] Kb = [HCO3-][OH-] = * = X2 [CO32-] X 1.79 * = X2 0.080-X 0.080 (1.79 * 10-4 ) = X2 X = = [OH-] :pOH- = 2.42 : pH =
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H3PO4(aq) H+ (aq) + H2PO4- (aq) H3PO4 H+ H2PO4-
MONSTER! 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. . H3PO4 H+ H2PO4- 0.025 M -X X 0.025-X Ka1 = [H+][H2PO4-] [H3PO4] 7.5 X 10-3 = [X][X] [0.025-X]
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The H2PO4- WILL DISSOCIATE IN Ka2
MONSTER! 7.5 X = [X][X] [0.025-X] Assume all H+ is from Ka1, it will appear in the ICE charts of Ka2, Ka3 7.5 X = (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= X (7.5 X 10-3 )2 - 4(1) ( x 10-4) : X= = [H+] 2(1) X= =[H2PO4- ] The H2PO4- WILL DISSOCIATE IN Ka2 H2PO4- H+ HPO42- -X X X X
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H2PO4-(aq) H+ (aq) + HPO42- (aq) H2PO4- H+ HPO42-
MONSTER! REACTION #2 : DISSOCIATION Ka2 OF POLY (TRI) PROTIC ACID. ALL INOZATIONS ARE WEAK. H2PO4-(aq) H+ (aq) + HPO42- (aq) RECOGNIZE THIS AS A WEAK POLYPROTIC ACID IONIZING IN PURE WATER. . Ka2 = [H+][HPO42-] [H2PO4-] H2PO4- H+ HPO42- -X X X 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]
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HPO42- H+ PO43- 6.2x10-8 0.01045 -X X 16.113 MONSTER!
Ka2 = [0.010+X][X] [0.010-X] Ka2 = 6.2 x 10-8 = 0.010X 0.010 X = 6.2 X 10-8 = [HPO42-], THE SPECIES OF INTEREST FROM Ka2 FOR Ka3 HPO42- H+ PO43- 6.2x10-8 -X X 6.2x10-8 -X X
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HPO42-(aq) H+ (aq) + PO43- (aq) HPO42- H+ PO43-
MONSTER! 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. . Ka2 = [H+][HPO42-] [H2PO4-] HPO42- H+ PO43- 6.2x10-8 0.010 -X X 6.2x10-8 -X X From Ka2 Ka2 = [0.010+X][X] [6.2x10-8-X] Ka3 = 4.2 x = 0.010X 6.2x10-8 X = 2.6 x = [PO43-], THE SPECIES OF INTEREST FROM Ka3
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