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Calculatethe [OH^(-)]of a solution after100 mLof 0.1M MgCl_(2)is addedto 100 mL0.2M NaOH K_(sp)of Mg(OH)_(2) is 1.2 xx10^(-11) . |
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Answer» Solution :` {:(,MgCl_(2),+,2NaOH,to,"Mg"(OH)_(2),+,2NaCl),(" mm before REACTION",10,,20,,0,,0),(,0,,0,,10,,20):}` Thus , 10 m moleof `Mg(OH)_(2)`are formes . The productof `[Mg^(2+)][OH^(-)]^(-2)` is therefore`[(100)/(200)]XX [(20)/(200)]^(2) = 5 xx10^(-4)`whichis more than`K_(SP)`of `Mg(OH)_(2)`, Now SOLUBILITY (S) of`Mg(OH)_(2)`can bederived by ` K_(sp) = 4S^(3)` ` :. S = root3(K_(sp)) = root3(1.2 xx 10^(-11)) = 2.29 xx10^(-4)` ` :. [OH^(-)] = 2S = 2xx 2.29 xx10^(-4) = 4.58 xx10^(-4)`. |
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