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(B) (BH + )

(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

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(B) (BH + )

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  1. (B) (BH+) C2H5NH2(aq) + H20   C2H5NH3+ (aq) + OH- (aq) 16.75 by forum request A conjugate acid of a weak base A weak base Kb = [C2H5NH3+ ] [OH- ] = (X) (X) = X2 = 6.4 x 10-4 [C2H5NH2] (0.075-X) (0.075-X)

  2. 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

  3. 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

  4. CN- + H2O  OH- + HCN HYDROLYSIS OF CN-, CONJUGATE BASE. 16.83 a 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

  5. 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

  6. 16.83 b 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

  7. 16.113 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. . Ka1 = [H+][H2PO4-] [H3PO4] 7.5 X 10-3 = [X][X] [0.025-X]

  8. 16.113 MONSTER! 7.5 X 10-3 = [X][X] [0.025-X] Assume all H+ is from Ka1, it will appear in the ICE charts of Ka2, Ka3 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-] The H2PO4- WILL DISSOCIATE IN Ka2

  9. 16.113 MONSTER! Ka2 = [H+][HPO42-] [H2PO4-] Ka2 = [0.010+X][X] [0.010-X] 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. . IGNORE ANY H+ PRODUCED BY Ka2, ASSUME ALL H+ IS FROM Ka1 Ka2 = [0.010+X][X] [0.010-X]

  10. 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-], THESPECIES OF INTEREST FROM Ka2 FOR Ka3

  11. 16.113 MONSTER! Ka2 = [H+][HPO42-] [H2PO4-] Ka2 = [0.010+X][X] [6.2x10-8-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 Ka3 = 4.2 x 10-13 = 0.010X 6.2x10-8 X = 2.6 x 10-18=[PO43-], THESPECIES OF INTEREST FROM Ka3

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