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Analyzing the Hydrolysis of Weak Bases and Conjugate Acids for pH Calculation

This document provides a comprehensive analysis of the hydrolysis reactions involving weak bases, their conjugate acids, and polyprotic acids. It details the equilibrium expressions, calculations for pH and pOH, and the behavior of salts derived from weak acids and strong bases. The techniques involve identifying Kb and Ka values, forming ICE tables to determine concentrations, and understanding the hydrolysis of ions in solution. The methodology is crucial for predicting the acidity or basicity of solutions and mastering acid-base chemistry.

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Analyzing the Hydrolysis of Weak Bases and Conjugate Acids for pH Calculation

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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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