Ch 7_Part 2_slide corrections_112811

# Ch 7_Part 2_slide corrections_112811 - From dissociation of...

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Calculate the pH of 3.00 x 10 -3 M Sulfuric Acid HSO 4 - (aq) SO 4 2- (aq) + H + (aq) K a2 = 1.2 x 10 -2 H 2 SO 4 (aq) HSO 4 - (aq) + H + (aq) K a1 = LARGE Polyprotic acid, more than one proton to lose!!! Multiple K a ’s Because first dissociation has a very large K a it can be treated as a strong acid (~100% dissociated).

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Calculate the pH of 3.00 x 10 -3 M Sulfuric Acid HSO 4 - (aq) SO 4 2- (aq) + H + (aq) K a = 1.2x10 -2 Initial Concentration (mol/L) Equilibrium Concentration (mol/L) [HSO 4 - ] 0 = 0.00300 [HSO 4 - ] = 0.00300 – x [SO 4 2- ] 0 = 0 [SO 4 2- ] = x [H + ] 0 = 0.00300 [H + ] = 0.00300 + x x mol/L HSO 4 - dissociates to reach equilibrium
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Unformatted text preview: From dissociation of H 2 SO 4 [H + ][SO 4 2-] [HSO 4-] K a2 = 1.2 x 10-2 = = (0.00300 + x)(x) (0.00300 – x) No obvious approximation looks valid, so we must solve with the quadratic formula. 0 = x 2 + 0.015x – 3.6 x 10-5 a = 1 b = 0.015 c = -3.6 x 10-5 x = -b + - b 2 – 4ac 2a x = 2.10 x 10-3 [H + ] = 0.00300 + x = 0.00510 pH = 2.29 Note: Change in [H+] does not violate huge K a for 1 st dissociation: [H 2 SO 4 ] = 0. Small and highly charged cations produce acidic solutions in water . Examples: Al 3+ , Fe 3+ , etc. ....
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Ch 7_Part 2_slide corrections_112811 - From dissociation of...

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