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

Question

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

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.

Monjur Alahi

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2025-01-04 04:42

1 Answers

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Salim Ahmed Kawsar · Accepted Answer · 2023-02-05T15:29:48+06:00

Salim Ahmed Kawsar

`"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`

5 views Answered 2023-02-05 15:29