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The pH of 0.001 M `Ba(OH)_(2)` solution will beA. 2B. 8.4C. 11.3D. 2.7 |
Answer» Correct Answer - C `Ba(OH)_(2)hArrBa^(2+)+2OH^(-)` `[OH^(-)]=2xx1xx10^(-3)M` `pOH=-log[OH^(-)]=-log(2xx10^(-3))=2.7` `pOH+pH=14pH=14-2.7=11.3` |
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