The `K _(sp ) of Ca (OH) _(2) is 4.42 xx 10 ^(-4) at 25 ^(@)C. A 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 is precipitated?

Options

A. `527.3 mg`
B. `638.4 mg`
C. `218.3 mg`
D. `743mg`

Answer

D

Explanation

500 mL of `0.4 M NaOH` are mixed with 500 mL of `Ca (OH) _(2)` a saturated solution having `Ca (OH) _(2)` solubility as SM
For `Ca (OH) _(2) hArr Ca ^(2+) + 2 OH ^(-) K _(sp ) = S xx (2S ) ^(2) = 4S ^(3)`
`therefore S = sqrt (( 4.42 xx 10 ^(-5))/( 4 )) = 0.0223 M`
Now `Ca (OH)_(2) + NaOH` are mixed `therefore `Solution has `Ca ^(2+) and OH ^(-)` out of which some `Ca ^(2+)` wil precipitate. On mixing `[Ca ^(2+) ] = ( 0. 0 223 xx 2 xx 500)/( 1000) = 0. 0 111 5 = 111. 5 xx 10 ^(-4) M`
`[OH ^(-)] = ( 0 .0223 xx 2 xx 500)/( 1000) + ( 500 xx 0. 4)/(1000) `[ from Ca `(OH) _(2) ]` [from`NaOH] = 0.223M`
`[Ca ^(2+)] [OH ^(-)] ^(2) = K _(sp), [Ca ^(2+)] _("left") [0.2223]^(2) = 4.42 xx 10 ^(-5)`
`[Ca ^(2+)] _("left") (4.42 xx 10 ^(-5))/( [ 0. 222 3] ^(2)) = 8. 9 4 xx 10 ^(-4)` mol/L
`therefore ` mole of `Ca (OH) _(2)` precipitated = Mole of `[Ca ^(2+)]`
precipitated `=111.5 xx 10 ^(-4) - 8. 94 xx 10 ^(-4) = 102. 46 xx 10 ^(-4)`
`therefore ` wt of `Ca (OH _(2))` precipitated `=102 . 46 xx 10 ^(-4) xx 74 = 7582.04 xx 10 ^(-4) g = 758 . 2 mg`