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If the concentration of `Mg^(2+)` ions in sea water is 1200 ppm. How many moles of NaOH are required to precipitate all `Mg^(2+)` ions into `Mg(OH)_(2)(S)` present in 1 litre solution. |
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Answer» `"ppm"=(W_(Mg^(+2)))/(W_("water"))xx10^(6)` `10^(6)ml` solution contain `W_(Mg^(+2))=1200`ppm `[Mg^(+2)]=(1200)/(24)xx(10^(3))/(10^(6))` `[Mg^(+2)]=50xx10^(-3)=5xx10^(-2)=n_(Mg^(+2))` in 1 litre solution `n_(NaOH)` used =`2xxn_(Mg^(+2))` `=2xx5xx10^(-2)` `=10^(-1)xx10=1` |
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