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In a solution of `0.04M FeCl_(2)` and `0.01 M FeCl_(3)`, how large may be its pH of without being precipitation of either `Fe(OH)_(2)` or `Fe(OH)_(3)` ? [Given `K_(sp) Fe(OH)_(2) = 16 xx 10^(-6)` and `K_(sp) Fe(OH)_(3) = 8 xx 10^(-26)`]A. `5.7`B. `6.3`C. `8.3`D. `10.7` |
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Answer» Correct Answer - B For the precipitation of `Fe(OH)_(2)` `K_(sp) = [0.04][OH^(-)]^(2) = 16 xx 10^(-6)` `[OH^(-)] = 2 xx 10^(-2)` and for `Fe[OH]_(3)` `K_(sp) = 0.02 xx [OH^(-)]^(2) = 16 xx 10^(-6)` `[OH^(-)] = 2 xx 10^(-2)` For `Fe(OH)_(3)` the `[OH^(-)]` ions are required less `pOH = 8 - "log" 2` `pH = 6 + "log" 2 = 6.3` |
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