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Calculate the [OH^(-)] of a solution a...

Calculate the `[OH^(-)]` of a solution after 100 mL of 0.1 M `MgCl_(2)` is added to 100 mL 0.2 M NaOH `K_(sp) ` of `Mg(OH)_(2)` is `1.2 xx10^(-11)` .

Text Solution

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` {:(,MgCl_(2),+,2NaOH,to,"Mg"(OH)_(2),+,2NaCl),(" mm before reaction",10,,20,,0,,0),(,0,,0,,10,,20):}`
Thus , 10 m mole of `Mg(OH)_(2)` are formes . The product of `[Mg^(2+)][OH^(-)]^(-2)` is therefore `[(100)/(200)] xx [(20)/(200)]^(2) = 5 xx10^(-4)` which is more than `K_(sp)` of `Mg(OH)_(2)` , Now solubility (S) of `Mg(OH)_(2)` can be derived 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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