A solution contains A13+ and Co2+. The addition of 0.3731 L of 1.735 M NaOH results in the complete precipitation of the ions as Al(OH), and Co(OH)2. The total mass of the precipitate is 22.73 g. Find the masses of Al3+ and Co2+ in the solution.

The correct answer is mass of Al3+ will be 3.23 grams and the mass of Co2+ will be 8.50 grams.

Explanation:

Based on the given information, 0.3731 L of 1.735 M of NaOH is added in a solution resulting in the precipitation of the ions as Al(OH)₃ and Co(OH)₂. Thus, the moles of NaOH will be molarity × V(L) = 1.735 × 0.3731 L = 0.647 moles.  

The mass of the precipitate given is 22.73 grams.  

Now let us assume that the mass of Al(OH)₃ will be x grams and the mass of Co(OH)₂ will be (22.73-x) grams

Therefore, the moles of Al(OH)₃ will be x grams/78 g/mol and as 3OH⁻ ions are needed so the moles will be 3x/78 mole.  

And, the moles of Co(OH)₂ will be (22.73-x)grams/92.94 g/mol and as 2OH⁻ ions are needed so the moles will be 45.46-2x/92.94 moles.

Now the equation will become,  

3x/78 + 45.46-2x/92.94 = 0.647 moles

0.03846 x + 0.489 - 0.02152 x = 0.647  

0.01694 x + 0.489 = 0.647

0.01694 x = 0.158

x = 0.158/0.01694

x = 9.327 grams

Hence, the mass of Al(OH)₃ is 9.327 grams, and the mass of Al³⁺ will be,  

= 9.327 gm/78 g/mol × 27 g/mol = 3.23 grams

Now the mass of Co(OH)₂ will be, (22.73 - 9.327) grams = 13.403 grams

the mass of Co²⁺ will be,  

= 13.403 grams / 92.94 g/mol × 58.94 g/mol = 8.50 grams

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