Saved Bookmarks
| 1. |
The solubility product of Ca(OH)_(2) at 250^(0)C is 4.42 xx 10^(-5). A 500 mL of saturated solution of Ca(OH)_(2) is mixed with equal volume of 0.4 M NaOH. How much Ca(OH)_(2) on milligram is precipitated? |
|
Answer» `758.2mg` For `Ca(OH)_(2) hArr Ca^(2+) + 2OH^(-)` `K_(sp) = s xx (2s)^(2)= 4s^(3)` Then, `4s^(3) = 4.42 xx 10^(-5)` `s=3sqrt(((4.42 xx 10^(-5))/(4))) = 0.0233M` Now `Ca(OH)_(2) + NaOH` are mixed `:.` Solution has `Ca^(2+)` and `OH^(-)` out of which some `Ca^(2+)` and `OH^(-)` out of which some `Ca^(2+)` are precipitates on mixing. `[Ca^(2+)] = (0.0223 xx 500)/(1000) = 0.01115 = 111.5 xx 10^(-4) M` `[OH^(-)] = (0.0223 xx 2 xx500)/(1000) + (500 xx 0.4)/(1000) = 0.2223 M` (from `Ca(OH)_(2)`), (from `NaOH`) `[Ca^(2+)] [OH^(-)]^(2) = KSP` `[Ca^(2+)]_("left") [0.2223]^(2)= 4.42 xx 10^(-5)` `[Ca^(2+)]_("left") = (4.42 xx 10^(-5))/([0.2223]^(2)) = 8.94 xx 10^(-4) "MOL litre"^(-1)` `:'` MOLE of `Ca(OH)_(2)` precipitated `=` Mole of `[Ca^(2+)] "INITIAL" - [Ca^(2+)]` final `= 111.5 xx 10^(-4) -8.94 xx 10^(-4)` `= 102.46 xx 10^(-4)` `:.` WEIGHT of `Ca(OH)_(2)` precipitated from `Ca(OH)_(2)` solution `= 102.46 xx 10^(-4) xx 74` `= 7582.04 xx 10^(-4) g = 758.2 mg` |
|