The reaction is given as:
[tex]2NaOH+Cd(NO_{3})_{2}\rightarrow Cd(OH)_{2}+NaNO_{3}[/tex]
Here, one mole of cadmium nitrate gives one mole of cadmium hydroxide.
Now,
number of moles = [tex]Concentration \times Volume[/tex] (1)
Concentration = 0.530 M
Volume = 35.0 mL
Substitute the above values in formula (1):
number of moles = [tex] 0.530 M\times 35.0 mL \times 10^{-3} L[/tex] (1 L =1000 mL)
= [tex]18.55\times 10^{-3} moles[/tex]
From the reaction, two moles of sodium hydroxide is needed to give one mole of cadmium hydroxide.
Now,
number of moles of sodium hydroxide which is needed to give [tex]18.55\times 10^{-3} moles[/tex] of cadmium hydroxide.
Number of moles of sodium hydroxide =[tex]18.55\times 10^{-3} moles of Cd(OH)_{2} \times \frac{2 mol of NaOH}{1 mol of Cd(OH)_{2}}[/tex]
= [tex]37.1\times 10^{-3} mol[/tex]
Now, mass of sodium hydroxide is calculated by:
Mass of sodium hydroxide =[tex]Moles\times Molar mass[/tex]
= [tex]37.1\times 10^{-3} mol\times 40 g/mol[/tex]
= [tex]1.484 g[/tex]
Thus, mass of sodium hydroxide required is [tex]1.484 g[/tex]
The mass of NaOH needed to precipitate the Cd2+ ions from 35.0 mL, 0.530 M Cd(NO3)2 solution would be 1.484 grams
From the equation of the reaction:
Cd(NO3)2 + 2 NaOH → Cd(OH)2 + 2 NaNO3
Mole ratio of Cd(NO3)2 and NaOH = 1:2
Mole of 35.0 mL, 0.530 M Cd(NO3)2 = 35/1000 x 0.530
= 0.0186 mole
Equivalent mole of NaOH = 0.0186 x 2 = 0.0371 moles
Mass of 0.0371 mole NaOH = 0.0371 x 40
= 1.484 grams
More on stoichiometric reactions can be found here: https://brainly.com/question/6907332
