Answer:
0.0238M SbCl3, 1.07M H+, 1.14M Cl-
Explanation:
The total volume of the solution is:
4mL + 5.00mL + 12.0mL = 21mL
As the volume of the SbCl3 is 5.00mL, the dilution factor is:
21mL / 5.00mL = 4.2 times
The concentration of SbCl3 is:
0.10M SbCl3 / 4.2 times = 0.0238M SbCl3
The concentration of H+ = [HCl]:
4.5M / 4.2 times = 1.07M H+
The initial concentration of Cl- is:
3 times SbCl3 + HCl = 0.10M*3 + 4.5M =
<em>3 times SbCl3 because 1 mole of SbCl3 contains 3 moles of Cl-</em>
4.8M Cl- / 4.2 times = 1.14M Cl-
