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Bromine in excess is dropped to a 0.01 M SO 2 . All of SO 2 is oxidized to H 2 SO 4 and the excess Br 2 is removed by flushing with gaseous N 2 . Determine the pH of the resulting solution assuming K a 1 of H 2 SO 4 very large & K a2 = 10 − 2 . Take the value of log ( 3 . 24 ) = 0 . 51 .

Correct answer

1.49

Step-by-step solution

HSO 4 − 0 .01-x aq ⇌ H + 0 .03+x aq + SO 4 2+ x ; Ka 2 = 10 − 2 Ka 2 = 10 − 2 = x 0.03 + x 0.01 − x x = 2.36 × 10 − 3 = 0.00236 H + Total = 0.03 + 0.00236 = 0.03236 = 3.24 + 0.00236 pH = − log 3.240 × 10 − 2 = 2 − log 3.24 = 1.49

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