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Which of 1 m h 2 so 4 aq and 0 1 m hbraq has a higher

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which of 0 . 1 M H 2 SO 4 (aq) and 0 . 1 M HBr(aq) has a higher pH? 1. They have the same strength. 2. HBr correct 3. H 2 SO 4 Explanation: Both ionize 100% but H 2 SO 4 also has an additional H + from the second ionization HSO - 4 H + + SO 2 - 4 which increases the concentration of H + and results in a lower pH. 017 3.3points What is the pH of a 0.020 M solution of hydrosulfuric acid, a diprotic acid? K a1 = 1 . 1 × 10 - 7 K a2 = 1 . 0 × 10 - 14 1. 9.67 2. 4.69 3. 3.65 4. 4.33 correct 5. 5.22 6. 7.00 7. 7.84 Explanation: Solve using ONLY the 1st ionization. So this works like any other monoprotic acid where the assumption [H + ] = radicalbig (Conc)( K a1 ) is valid. 018 3.3points The hypothetical weak acid H 2 A ionizes as shown below. H 2 A + H 2 O H 3 O + + HA - K 1 = 1 × 10 - 7 HA - + H 2 O H 3 O + + A 2 - K 2 = 5 × 10 - 11 Calculate the [HA - ] in a 0.20 M solution of H 2 A. 1. 1 . 4 × 10 - 4 M correct 2. 1 . 0 × 10 - 7 M 3. 3 . 0 × 10 - 4 M 4. 6 . 3 × 10 - 5 M 5. 2 . 2 × 10 - 6 M Explanation: [H 2 ] ini = 0.20 M Let x = [HA - ]. Since we’re considering a solution of H 2 A and HA - , we should use the expression for K 1 here: K 1 = [H + ] [HA - ] [H 2 A] 1 × 10 - 7 = ( x ) ( x ) 0 . 2 - x

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casey (rmc2555) – Homework 8 – holcombe – (51395) 6 As H 2 A has a very small K 1 , we can assume that x is very small compared to 0 . 2 and simplify our equation to 1 × 10 - 7 = ( x ) 2 (0 . 2) x 2 = 2 × 10 - 8 x = 0 . 000141421 Thus x = [HA - ] = 0 . 000141421 M. 019 3.3points What would be the pH of a 0 . 39 M solution of H 2 SO 4 ? H 2 SO 4 has a K a1 that is so large we can assume complete dissociation and a K a2 of 0 . 012. 1. 0 . 409 2. 0 . 397 correct 3. 0 . 0951 4. 0 . 108 Explanation: Because H 2 SO 4 is a strong acid, the re- action H 2 SO 4 H + + HSO 4 - goes to com- pletion. In this case, producing a [H + ] of 0 . 39 M . Because HSO 4 - is a weak acid we must solve for the proton concentration it contributes to the total. HSO 4 - ←→ H + SO 4 2 - 0 . 39 0 . 39 0 - x + x + x 0 . 39 - x 0 . 39 + x x K a = (0 . 39 + x ) x 0 . 39 - x 1 . 2 × 10 - 2 = (0 . 39 + x ) x 0 . 39 - x x 2 + 0 . 402 x = - 0 . 00468 x = 0 . 0113229 [H + ] = 0 . 39 M pH = 0 . 397 020(part1of2)3.3points An aqueous solution has a dilute HNO 3 con- centration of 1 . 0 × 10 - 10 M at 298 K. What is the greatest contributor toward solution pH? 1. autoionization of water correct 2. concentration of the acid HNO 3 3. temperature 4. extend of ionization of the acid HNO 3 Explanation: The strong acid is 10 3 times more dilute than the H + in solution due to the water (solvent). The pH depends mostly on the autoionization of water in this case. 021(part2of2)3.3points What is the pH of the above solution? 1. 10 2. 4 3. 7 correct 4. 1 Explanation: At 298 K, the [H + ] due to autoionization of water is 1 × 10 - 7 , so the pH = 7. 022 3.3points What is the pH of 2 × 10 - 9 M Ba(OH) 2 ? 1. 7.02 correct 2. 5.30 3. 8.40 4. 8.70 5. 5.60 Explanation: [Ba(OH) 2 ] = 2 × 10 - 9 M K w = 1 × 10 - 14 Ba(OH) 2 completely dissociates: Ba(OH) 2 (aq) Ba 2+ (aq) + 2 OH - (aq)
casey (rmc2555) – Homework 8 – holcombe – (51395) 7 [OH - ] = 2 [Ba(OH) 2 ] = 2 (2 × 10 - 9 ) = 4 × 10 - 9 which is less than the [OH - ] in pure water (1 × 10 - 7 ), so we must consider this concen- tration: H 2 O H + + OH - ini - 1 × 10 - 7 1 × 10 - 7 Δ - - 4 × 10 - 9 +4 × 10 - 9 fin 9 . 6 × 10 - 8 1 . 04 × 10 - 7 K w = [OH - ] [H + ] [H + ] = K w [OH - ] = 1 × 10 - 14 1 . 04 × 10 - 7 = 9 . 61538 × 10 - 8 Thus pH = - log[H + ] = - log(9 . 61538 × 10 - 8 ) = 7 . 01703 Remember to check if your pH makes sense.

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