Answer:
0.0714 M for the given variables
Explanation:
The question is missing some data, but one of the original questions regarding this problem provides the following data:
Mass of copper(II) acetate: ![m_{(AcO)_2Cu} = 0.972 g](https://tex.z-dn.net/?f=m_%7B%28AcO%29_2Cu%7D%20%3D%200.972%20g)
Volume of the sodium chromate solution: ![V_{Na_2CrO_4} = 150.0 mL](https://tex.z-dn.net/?f=V_%7BNa_2CrO_4%7D%20%3D%20150.0%20mL)
Molarity of the sodium chromate solution: ![c_{Na_2CrO_4} = 0.0400 M](https://tex.z-dn.net/?f=c_%7BNa_2CrO_4%7D%20%3D%200.0400%20M)
Now, when copper(II) acetate reacts with sodium chromate, an insoluble copper(II) chromate is formed:
![(CH_3COO)_2Cu (aq) + Na_2CrO_4 (aq)\rightarrow 2 CH_3COONa (aq) + CuCrO_4 (s)](https://tex.z-dn.net/?f=%28CH_3COO%29_2Cu%20%28aq%29%20%2B%20Na_2CrO_4%20%28aq%29%5Crightarrow%202%20CH_3COONa%20%28aq%29%20%2B%20CuCrO_4%20%28s%29)
Find moles of each reactant. or copper(II) acetate, divide its mass by the molar mass:
![n_{(AcO)_2Cu} = \frac{0.972 g}{181.63 g/mol} = 0.0053515 mol](https://tex.z-dn.net/?f=n_%7B%28AcO%29_2Cu%7D%20%3D%20%5Cfrac%7B0.972%20g%7D%7B181.63%20g%2Fmol%7D%20%3D%200.0053515%20mol)
Moles of the sodium chromate solution would be found by multiplying its volume by molarity:
![n_{Na_2CrO_4} = 0.0400 M\cdot 0.1500 L = 0.00600 mol](https://tex.z-dn.net/?f=n_%7BNa_2CrO_4%7D%20%3D%200.0400%20M%5Ccdot%200.1500%20L%20%3D%200.00600%20mol)
Find the limiting reactant. Notice that stoichiometry of this reaction is 1 : 1, so we can compare moles directly. Moles of copper(II) acetate are lower than moles of sodium chromate, so copper(II) acetate is our limiting reactant.
Write the net ionic equation for this reaction:
![Cu^{2+} (aq) + CrO_4^{2-} (aq)\rightarrow CuCrO_4 (s)](https://tex.z-dn.net/?f=Cu%5E%7B2%2B%7D%20%28aq%29%20%2B%20CrO_4%5E%7B2-%7D%20%28aq%29%5Crightarrow%20CuCrO_4%20%28s%29)
Notice that acetate is the ion spectator. This means it doesn't react, its moles throughout reaction stay the same. We started with:
![n_{(AcO)_2Cu} = 0.0053515 mol](https://tex.z-dn.net/?f=n_%7B%28AcO%29_2Cu%7D%20%3D%200.0053515%20mol)
According to stoichiometry, 1 unit of copper(II) acetate has 2 units of acetate, so moles of acetate are equal to:
![n_{AcO^-} = 2\cdot 0.0053515 mol = 0.010703 mol](https://tex.z-dn.net/?f=n_%7BAcO%5E-%7D%20%3D%202%5Ccdot%200.0053515%20mol%20%3D%200.010703%20mol)
The total volume of this solution doesn't change, so dividing moles of acetate by this volume will yield the molarity of acetate:
![c_{AcO^-} = \frac{0.010703 mol}{0.1500 L} = 0.0714 M](https://tex.z-dn.net/?f=c_%7BAcO%5E-%7D%20%3D%20%5Cfrac%7B0.010703%20mol%7D%7B0.1500%20L%7D%20%3D%200.0714%20M)