Question

1. in run 1, you mix 8.1 mL of the 43 g/L MO solution (MO molar...

1. in run 1, you mix 8.1 mL of the 43 g/L MO solution (MO molar mass is 327.33 g/mol), 2.89 mL of the 0.040 M SnCl22 in 2 M HCl solution, 5.46 mL of 2.0M HCl solution, and 3.47 mL of 2.0M NaCl solution.

What is the [MO]?

a. In run 1, you mix 8.1 mL of the 43 g/L MO solution (MO molar mass is 327.33 g/mol), 2.89 mL of the 0.040 M SnCl22 in 2.0 M HCl solution, 5.46 mL of 2.0 M HCl solution, and 3.47 mL of 2.0M NaCl solution.

What is the [H33O++]? Remember that there is a contribution of H33O++ from two solutions.

b. In run 1, you mix 8.1 mL of the 43 g/L MO solution (MO molar mass is 327.33 g/mol), 2.89 mL of the 0.040 M SnCl22 in 2 M HCl solution, 5.46 mL of 2.0M HCl solution, and 3.47 mL of 2.0M NaCl solution.

What is the [Sn2+2+]?

c. In the previous part, you calculated the concentrations of the components of the reaction mixture for Run 1. Pretend that you also calculated the concentrations for Run 2 and Run 3 and you plot log([Sn2+2+]) vs log(kexptexpt). The slope of the best fit line is 1. What is the likely exponent for [Sn2+2+]?

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Answer #1

Mass concentration of MO = 43 g/L

volume of MO = 8.1 mL = 8.1 x 10-3 L

mass of MO = (Mass concentration of MO) * (volume of MO)

mass of MO = (43 g/L) * (8.1 x 10-3 L)

mass of MO = 0.3483 g

moles of MO = (mass of MO) / (molar mass of MO)

moles of MO = (0.3483 g) / (327.33 g/mol)

moles of MO = 1.06 x 10-3 mol

total volume = 8.1 mL + 2.89 mL + 5.46 mL + 3.47 mL = 19.92 mL = 19.92 x 10-3 L

[MO] = (moles of MO) / (total volume)

[MO] = (1.06 x 10-3 mol) / (19.92 x 10-3 L)

[MO] = 0.053 M

(a) moles of H3O+ = (molarity HCl solution) * (volume HCl solution)

moles of H3O+ = [(2.0 M) * (2.89 mL)] + [(2.0 M) * (5.46 mL)]

moles of H3O+ = 16.7 mmol

[H3O+] = (moles of H3O+) / (total volume)

[H3O+] = (16.7 mmol) / (19.92 mL)

[H3O+] = 0.84 M

(b) moles Sn2+ = (molarity SnCl2) * (volume SnCl2 solution)

moles Sn2+ = (0.040 M) * (2.89 mL)

moles Sn2+ = 0.1156 mmol

[Sn2+] = (moles Sn2+) / (total volume)

[Sn2+] = (0.1156 mmol) / (19.92 mL)

[Sn2+] = 0.0058 M

(c) The likely exponent for [Sn2+] is 1 (one)

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