What mass of NaOH
is needed to precipitate the Cd2+
ions from 39.0 mL
of 0.550 M
Cd(NO3)2
solution?
To find the mass of NaOH needed to precipitate the Cd2+ ions, we first need to determine the chemical reaction that occurs between NaOH and Cd(NO3)2.
The balanced chemical equation for the reaction between NaOH and Cd(NO3)2 is:
Cd(NO3)2 + 2 NaOH -> Cd(OH)2 + 2 NaNO3
From the balanced equation, we can see that 1 mole of Cd(NO3)2 reacts with 2 moles of NaOH.
To determine the amount of Cd(NO3)2 present in the solution, we use the formula:
moles = Molarity * Volume (L)
moles of Cd(NO3)2 = 0.550 mol/L * 0.0390 L = 0.02145 moles
From the chemical equation, we see that 1 mole of Cd(NO3)2 reacts with 2 moles of NaOH. Therefore, the moles of NaOH needed is:
moles of NaOH = 2 * 0.02145 moles = 0.0429 moles
Now, we can convert moles of NaOH to grams using the molar mass of NaOH:
Molar mass of NaOH = 22.99 g/mol (Na) + 15.999 g/mol (O) + 1.008 g/mol (H) = 39.997 g/mol
Mass of NaOH = moles of NaOH * molar mass of NaOH
Mass of NaOH = 0.0429 moles * 39.997 g/mol = 1.719 g
Therefore, 1.719 grams of NaOH is needed to precipitate the Cd2+ ions from 39.0 mL of 0.550 M Cd(NO3)2 solution.