[Cr(H2O)6]Cl3
[Cr(H2O)5Cl]H2O·Cl2
[Cr(H2O)4Cl2]Cl·2H2O
[Cr(H2O)3Cl3]3H2O
Moles of NaOH used = 0.0285 L × 0.125 mol/L = 0.0035625 mol
Since NaOH is a strong base, it reacts completely with the acid. The number of moles of H+ ions released is equal to the moles of NaOH used. Therefore, 0.0035625 moles of H+ ions were released.
Moles of CrCl3·6H2O = 0.319 g / 266.5 g/mol = 0.001197 mol
Let's consider the possible complexes and their dissociation in the cation exchange resin:
A. [Cr(H2O)6]Cl3 → [Cr(H2O)6]³⁺ + 3Cl⁻
This complex releases 3 moles of H⁺ per mole of complex due to the cation exchange.
B. [Cr(H2O)5Cl]Cl2·H2O → [Cr(H2O)5Cl]²⁺ + 2Cl⁻ + H2O
This complex releases 2 moles of H⁺ per mole of complex.
C. [Cr(H2O)4Cl2]Cl·2H2O → [Cr(H2O)4Cl2]⁺ + Cl⁻ + 2H2O
This complex releases 1 mole of H⁺ per mole of complex.
D. [Cr(H2O)3Cl3]3H2O → [Cr(H2O)3Cl3] + 3H2O
This complex does not release any H⁺.
Now we calculate the number of moles of H+ released for each complex given the moles of CrCl3·6H2O:
A. Moles of H+ = 3 × 0.001197 mol = 0.003591 mol
B. Moles of H+ = 2 × 0.001197 mol = 0.002394 mol
C. Moles of H+ = 1 × 0.001197 mol = 0.001197 mol
Comparing the calculated moles of H+ with the experimental value (0.0035625 mol), we find that complex A ([Cr(H2O)6]Cl3) is the closest. The small difference can be attributed to experimental error.
Therefore, the correct formula of the complex is [Cr(H2O)6]Cl3.