Question

Part A of the lab involved adding 4 mL increments of distilled water to 5.00 mL of antimony trichloride solution. The antimony trichloride solution contains 0.10 M SbCl3 in 4.5 M HCl. Calculate the concentrations of SbCl3 and H /Cl- in the test tube after 12.0 mL of distilled water has been added. Assume dilution only.

Answer 1

0.0238M SbCl3, 1.07M H+, 1.14M Cl-

Step-by-step 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 =

3 times SbCl3 because 1 mole of SbCl3 contains 3 moles of Cl-

4.8M Cl- / 4.2 times = 1.14M Cl-