© All Rights reserved @ LearnWithDash
Step 1: Write the Balanced Chemical Equation
The complex salt reacts with silver nitrate to precipitate chloride ions. The balanced reaction is:
[Co(NH_3)_6]Cl_3 + 3\,AgNO_3 \rightarrow 3\,AgCl + [Co(NH_3)_6](NO_3)_3
Step 2: Calculate Moles of the Complex Salt
Given mass of [Co(NH_3)_6]Cl_3 = 0.3\,\text{g} and its molar mass is 267.46\,\text{g/mol} . The number of moles is:
\text{Moles of }[Co(NH_3)_6]Cl_3
= \dfrac{0.3\,\text{g}}{267.46\,\text{g/mol}}
= 1.121\times 10^{-3}\,\text{mol}\ (\text{approximately}).
Step 3: Relate Moles of Complex Salt to Moles of AgNO₃
From the balanced equation, 1 mole of [Co(NH_3)_6]Cl_3 requires 3 moles of AgNO_3 to precipitate its three chloride ions. Therefore:
\text{Moles of }AgNO_3
= 3 \times \text{(Moles of }[Co(NH_3)_6]Cl_3)
= 3 \times 1.121\times 10^{-3}
= 3.363\times 10^{-3}\,\text{mol}.
Step 4: Calculate Volume of AgNO₃ Solution
The molarity of AgNO_3 solution is 0.125\,\text{M} (i.e., 0.125\,\text{mol/L} ). Using the relation:
\text{Moles of solute}
= \text{Molarity} \times \text{Volume (in L)}
3.363\times 10^{-3}\,\text{mol}
= 0.125\,\text{mol/L} \times V_{\text{(in L)}}
So,
V_{\text{(in L)}}
= \dfrac{3.363\times 10^{-3}}{0.125}
= 2.6904\times 10^{-2}\,\text{L}.
Step 5: Convert Volume to mL
1\,\text{L} = 1000\,\text{mL}.
Therefore,
V_{\text{(in mL)}}
= 2.6904\times 10^{-2} \times 1000
= 26.90\,\text{mL (approximately)}.
Final Answer
The volume of 0.125\,\text{M} AgNO_3 required is approximately 26.9 mL.
