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The K_(sp)of Ca(OH)_(2)is 4.42xx10^(-5)at 25^(@)C. 500 mL of saturated solution of Ca(OH)_(2) is mixed with equal volume of 0.4 M NaOH . How much Ca(OH)_(2)in mg will be precipitated ? Also calculate percentage precipitation. |
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Answer» Solution :Suppose the solubility of `Ca(OH)_(2)` in saturated solution is S mol `L^(-1)`. `{:(Ca(OH)_(2),hArr,Ca^(2+),+,2OH^(-),),(,,S,,2S,):}` `K_(sp) ` of `Ca(OH)_(2) hArr [Ca^(2+) ] [ OH^(-)]^(2)` `4.42xx10^(-5) = S xx (2S)^(2) = 4S^(3)` or`S=((4.42xx10^(-5))/(4))^(1//3) = 0.223 ` mol `L^(-1)` After mixing the two solutions, total volume becomes = 500 + 500 = 100 mL Now, concentration will be `[Ca^(2+) ] = (0.0223)/(1000) xx 500 = 0.01115 ` mol `L^(-1)` `{:([OH^(-)]=(0.0223xx2xx500)/(1000)+(0.4xx500)/(1000)=0.2223 " mol "L^(-1)),(""["From"Ca(OH)_(2)]"" ["From" NAOH]):}` On ADDING NaOH solution to saturated solution of `Ca(OH)_(2)` some `Ca(OH)_(2)` will be precipitated (DUE to common ion effect). Now, as `Ca^(2+)` ion left are still present in the left out saturated solution, `[ Ca^(2+) ] _("left") [OH^(-)]^(2) = K_(sp)` `:. [Ca^(2+) ]_("left") = (K_(sp))/([OH^(-)]^(2))=(4.42xx10^(-5))/((0.2223)^(2))=8.94xx10^(-4)` mol `L^(-1)` Moles of `Ca(OH)_(2)` precipitated = Moles of `Ca^(2+)` ions precipitated `= [Ca^(2+) ] _("initial")-[Ca^(2+)]_("left")` `=0.01115-8.94xx10^(-4) = 111.5xx10^(-4) - 8.94xx10^(-4) = 102.56xx10^(-4)` moles `=102.56xx10^(-4) xx 74 g = 7589.44 xx 10^(-4) g = 758.944 ` mg Initial `[Ca(OH)_(2)] = 0.01115 ` mol `L^(-1) = 111.5xx10^(-4) ` mol `L^(-1)` `Ca(OH)_(2)` precipitated `= 102.56xx10^(-4) ` mol `L^(-1)` `:. ` % `Ca(OH)_(2)` precipitated `= (Ca(OH)_(2)"precipitated")/("Initial conc. ")xx100=(102.56xx10^(-4))/(111.5xx10^(-4))xx100=91.98% = 92 %` Note. The above example illustrates the method of calculation of amount precipitated and amount left after precipitation when to a saturated solution of a sparingly soluble salt present in excess, some solution containing a common ion is added . |
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