Ni(II) coordination compound in an octahedral arrangement. Determine exact formula of compound. Info given: 4.37x10-4 moles of complex in 425 mL of water. You find that 17.49 mL of .1000 M HCl is need to titrate.
To determine the exact formula of the Ni(II) coordination compound in an octahedral arrangement, we can use the information provided about the titration with HCl.
First, we need to determine the moles of HCl used in the titration:
Moles of HCl = Volume of HCl x Molarity of HCl
Moles of HCl = 17.49 mL * 0.1000 mol/L
Moles of HCl = 0.001749 mol
Since HCl reacts in a 1:1 ratio with the complex compound, the moles of HCl used in the titration are equal to the moles of complex present in the solution.
Moles of complex = 0.001749 mol
We also know the concentration and volume of the complex solution:
Concentration of complex solution = (4.37 * 10^-4 mol) / (425 mL)
Concentration of complex solution = 1.028 * 10^-6 mol/mL
Now, we can convert the concentration of the complex solution to moles of complex by multiplying it by the volume of water used in the titration (since the complex compound is dissolved in water):
Moles of complex = Concentration of complex solution x Volume of water
Moles of complex = (1.028 * 10^-6 mol/mL) * (425 mL)
Moles of complex = 4.367 * 10^-4 mol
Comparing this result to the moles obtained from the titration, we can see that they are very close.
Now, since the complex compound has a coordination number of 6 (octahedral arrangement), we know that there are six ligands coordinated to the central Ni(II) ion.
Therefore, the exact formula of the compound can be written as [NiL₆], where L represents the ligands coordinated to the Ni(II) ion.
Note: The specific identity of the ligands is not provided in the given information, so the exact formula of the compound would require additional information about the ligands coordinated to the Ni(II) ion.