What is the volume of 0.100 M sodium hydroxide needed to precipitate all of the nickel(II) ions from 150.0 mL of 0.250 M nickel(II) nitrate?

1 Answer
Aug 21, 2016

Answer:

We need a volume of #0.75*L#

Explanation:

#Ni^(2+) + 2OH^(-) rarr Ni(OH)_2(s)darr#

Alternatively,

#Ni(NO_3)_2(aq) + 2NaOH(aq) rarr Ni(OH)_2(s) + 2NaNO_3(aq)#

We need a stoichiometrically balanced equation to inform our calculations.

#"Moles of nickel ion"# #=# #150.0xx10^-3Lxx0.250*mol*L^-1=0.0375*mol#.

Because of the stoichiometry, we need #2xx0.0375*mol" hydroxide ion"=0.0750*mol#.

And thus volume of #0.100*mol*L^-1# #NaOH# #=# #(0.0750*mol)/(0.100*mol*L^-1)xx10^3*mL*L^-1# #=# #750*mL#