linglearning.com Strong Acias/Bases Chempen....D 3/26/2018 06:00 PM 52.2/100 3/2

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linglearning.com Strong Acias/Bases Chempen....D 3/26/2018 06:00 PM 52.2/100 3/21/2018 10:38 PM -Print Calculator Periodic Table uestion 12 of 32 Sapling Learning What are the concentrations of OH' and H' in a 0.00086 M solution of Ba(OH)2 at 25 ? Assume complete dissociation Number [OH -I5.81 × 10 12 IMI Number H- 101 There is a hint availablel View the hint by clicking on the bottom dvide divider bar again to hide the hint Close O Previous Give Up & View Soldan Check Answer 0 Nest Ext search

Explanation / Answer

Ba(OH)2 dissociates completely to form Ba2+ and 2OH-

[OH-] = 2 * 0.00086 = 0.00172 M

Therefore, [OH-] = 0.00172 M

[H+][OH-] = 1*10^-14

[H+] * 0.00172 = 1*10^-14

[H+] = (1*10^-14) / 0.00172 = 5.81*10^-12

Therefore, [H+] = 5.81*10^-12

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