Question

What mass of NaOH is needed to precipitate the Cd2+ ions from 40.0 mL of 0.510 M Cd(NO3)2 solution?

Answer 1

Answer 2

To determine the mass of NaOH needed to precipitate Cd2+ ions from the given Cd(NO3)2 solution, we need to calculate the moles of Cd2+ ions present and then determine the stoichiometric ratio between Cd2+ ions and NaOH.

First, let's calculate the moles of Cd2+ ions present in the solution: moles of Cd2+ = volume (in L) × concentration (in mol/L) = 0.035 L × 0.530 mol/L = 0.01855 mol Cd2+

From the balanced chemical equation between Cd2+ and NaOH, we know that one mole of Cd2+ reacts with two moles of NaOH to form Cd(OH)2 precipitate.

Therefore, the moles of NaOH required to precipitate Cd2+ ions are: moles of NaOH = 2 × moles of Cd2+ = 2 × 0.01855 mol = 0.03710 mol NaOH

To calculate the mass of NaOH needed, we can use the molar mass of NaOH, which is 22.99 g/mol (Na) + 16.00 g/mol (O) + 1.01 g/mol (H) = 39.00 g/mol.

mass of NaOH = moles of NaOH × molar mass of NaOH = 0.03710 mol × 39.00 g/mol = 1.4467 g

Therefore, approximately 1.45 grams of NaOH is needed to precipitate the Cd2+ ions from 35.0 mL of 0.530 M Cd(NO3)2 solution.