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What mass of NaOH is needed to precipitate the Cd2+ ions from 38.0mL of 0.520 M Cd(NO3)2 solution?

User Betohaku
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1 Answer

4 votes

Answer:

1.58 g NaOH

Step-by-step explanation:

First we need to find the moles of Cd2+ ions.

Since we know the concentration which is .520 moles/L we can multiply this by the volume to get moles of Cd(NO3)2. We must also divide the volume by 1000 to get it into L from mL.

0.520 moles / L x (0.0380 L) = 0.0198 moles Cd(NO3)2. Since there is 1 mole of Cd for every mole of Cd(NO3)2 there is 0.0198 moles of Cd.

Precipitate formed will be Cd(OH)2 (s). For every mole formed it requires 2 moles of OH. Therefore the moles of OH must be 0.0198 x 2 = 0.0395 moles OH.

For every mole of OH there is 1 Mole of NaOH. Molar Mass of NaOH = 22.99 + 16.00 + 1.008 = 40.00 g/mol

Therefore multiply the moles of OH by the molar mass.

0.0395 moles x 40.00 g/mol = 1.58 g of NaOH

User Normajean
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