Final answer:
To determine the masses of Al³⁺ and Co²⁺ ions, we use the volume and molarity of NaOH to calculate the moles of OH⁻ and distribute them according to stoichiometric ratios. These moles are then converted to masses using the molar masses of Al(OH)3 and Co(OH)2 to find the individual ion masses in the precipitate.
Step-by-step explanation:
To find the masses of Al³⁺ and Co²⁺ ions precipitated as Al(OH)3 and Co(OH)2, we use stoichiometry and the molar masses of the respective hydroxides. The precipitation reaction for aluminum is:
Al³⁺ + 3OH⁻ → Al(OH)3 (s)
For cobalt:
Co²⁺ + 2OH⁻ → Co(OH)2 (s)
Given that 0.3706 L of 1.723 M NaOH (which supplies OH⁻) fully precipitates the ions and the total mass of the precipitate is 22.63 g, we calculate the moles of OH⁻ supplied:
0.3706 L × 1.723 mol/L = moles of OH⁻
We split the moles of OH⁻ between Al(OH)3 and Co(OH)2 based on their stoichiometric ratios. The moles of Al(OH)3 and Co(OH)2 are then converted to grams using their respective molar masses, which gives the masses of precipitated Al³⁺ and Co²⁺ ions. This is a stoichiometry problem involving chemical precipitation.