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musickatia [10]
3 years ago
9

Determine the overall charge on each complex.a) tetrachlorocuprate(i)b) pentaamminechlorocobalt(iii)c) diaquadichloroethylenedia

minecobalt(iii)
Chemistry
1 answer:
balu736 [363]3 years ago
4 0
Complex a: <span>tetrachlorocuprate(i)
In present complex chlorine is negatively charged ligand (-1) and oxidation state of copper is +1. Therefore, total charge on complex = 4(-1) + 1 = -3

Complex b: </span><span>pentaamminechlorocobalt(iii)
In present complex, ammine is a neutral ligand (charge = 0), chlorine is negatively charged ligand (charge = -1) and oxidation state of Co is +3. Therefore, total charge on complex is 5(0) + (-1) + (+3) = +2

</span><span>Complex c: diaquadichloroethylenediaminecobalt(iii)
</span>In present complex, aqua is a neutral ligand (charge = 0), chlorine is a negatively charged ligand (charge = -1), ethylenediamine is a neutral ligand (charge = 0) and oxidation state of cobalt is +3. Therefore, total charge on complex is 2(0) + 2(-1) + 2(0) + 3 = +1. 
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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

Volume of the sodium chromate solution: V_{Na_2CrO_4} = 150.0 mL

Molarity of the sodium chromate solution: c_{Na_2CrO_4} = 0.0400 M

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)

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

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

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Write the net ionic equation for this reaction:

Cu^{2+} (aq) + CrO_4^{2-} (aq)\rightarrow CuCrO_4 (s)

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

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

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

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