Final answer:
The equivalent weight of the nickel complex containing one mole of NH₃ is 82.45 g/mol. To propose the molecular formula of the nickel salt, additional data on the salt's molar mass and the stoichiometry of the metal complex would be necessary.
Step-by-step explanation:
Equivalent Weight Calculation
To calculate the equivalent weight of the sample containing one mole of NH₃, we must first determine the moles of HCl and NaOH used in titration. The given information states that 30.00 mL of 0.1013 N HCl was used, and the excess required 6.30 mL of 0.1262 N NaOH to neutralize. Let's conduct the calculations step by step.
Moles of HCl initially added:
0.03000 L × 0.1013 mol/L = 0.003039 mol
Moles of NaOH used to neutralize the excess HCl:
6.30 mL × 0.1262 mol/L = 0.00079446 mol
Since NaOH and HCl react in a 1:1 ratio, the moles of NaOH indicate the leftover moles of HCl after reaction with the nickel compound.
Therefore, moles of HCl reacted with the nickel compound are:
0.003039 mol - 0.00079446 mol = 0.00224454 mol
Now, we know that this number of moles reacted with one mole of NH₃. So the equivalent weight of the sample is:
Equivalent weight = mass of sample / moles of NH₃ reacted
0.185 g / 0.00224454 mol = 82.45 g/mol
Propose a Molecular Formula
Now, knowing the equivalent weight for one mole of NH₃, we can deduce a potential formula for the nickel complex. Assuming Ni(NH₃)m(H2O)p is the general formula and knowing one mole of NH₃ reacts, if the molecular weight of the salt was equal to the equivalent weight, the compound would contain one NH₃ and the rest would be Ni and H2O. Since the equivalent weight we calculated is way above the molar mass of one NH₃, however, we must assume the presence of additional NH₃ molecules or water in coordination with the nickel ion. We would need additional data to calculate the exact molecular formula.