`1.75g` of solid `NaOH` is added to `0.25dm^(3)` of `0.1M NiCI_(2)` solution. Calculate:
a. Mass of `Ni(OH)_(2)` forms
b. `pH` if final solution Given `K_(sp)` of `Ni(OH)_(2) = 1.6 xx 10^(-14)`

Explanation

a. `{:(,2NaOH+,NiCI_(2)rarr,Ni(OH)_(2)+,2NaCI),("Mole before reaction",(1.75)/(40),0.25xx0.1,,),(,=0.0438,=0.025,,),("Mole left",0,(0.025-(0.0438)/(2)),(0.0438)/(2),0.0438):}`
`:.` Mole of `Ni(OH)_(2)` formed `= (0.0438)/(2)`
`:. w.t.` of `Ni(OH)_(2)` solution `= (0.0438)/(2) xx 92.6 = 2.0279 g`
b. Also for `Ni(OH)_(2)` solution,
`Ni(OH)_(2) hArr Ni^(2+) +2OH^(Theta)`
`[Ni^(2+)] [OH^(Theta)]^(2) = K_(sp)`
`[(0.05-(0.0438)/(2))/0.25][OH^(Theta)]^(2) = 1.6 xx 10^(-14)`
`:. [overset(Theta)OH] = 11.35 xx 10^(-7) :. pOH = 5.64 :. pH = 8.06`
Note: `Ni(OH)_(2)` formed will be precipitated out since maximum solubility of `Ni(OH)_(2)` is `1.58 xx 10^(-5)M`.