What mass of NaOH is needed to precipitate the Cd2 ions from

What mass of NaOH is needed to precipitate the Cd2+ ions from 36.0 mL of 0.490 M Cd(NO3)2solution?

Solution

The reaction is given as :

2NaOH + Cd(NO3)2 = Cd(OH)2 + 2NaNO3

Number of moles of Cd2+ ions = 0.036 x 0.490

= 0.01764 mol

So the number of mol of NaOH required = 2 x 0.01764

= 0.03528 mol

Mass of NaOH = 40 x 0.03528

= 1.41 grams