Question

In the laboratory a student combines 29.3 mL of a 0.176 M zinc chloride solution with 11.2 mL of a 0.362 M zinc acetate solution. What is the final concentration of zinc cation ? 68.9 M Submit Answer Retry Entire Group 8 more group attempts remaining

Answer 1

Solution :

1)  Zinc chloride and Zinc acetate are dissociated as,

ZnCl2 = Zn2+ + 2Cl-

Zn(CH3COO)2 = Zn2+ + 2CH3COO-

Thus,

Number of moles of Zn^2+ ion is same as ZnCl2 and Zn(CH3COO)2.

>>>>Number of milimoles of Zinc chloride

= Molarity x Volume = 0.176 M x 29.3 mL = 5.1568 milimol

>>>Number of milimoles of Zinc acetate

= Molarity x Volume = 0.362 M x 11.2 mL = 4.0544 milimol

Total milimoles of Zn2+ = 5.1568 + 4 0544 = 9.2112 milimol

Total volume = 29.3 mL + 11.2 mL = 40.5 mL

Thus,

Molarity of Zn2+ ion = Total milimoles / Total volume

= 9.2112 mmol / 40.5 mL = 0.2274 M

2) Number of molecules of Zinc iodide

= 24.4 mL x 0.488 M = 11.9072 milimol

Hence, milimoles of Iodide ion = 2 x 11.9072 = 23.8144 mol

Number of molecules of Aluminium iodide

= 19.1 mL x 0.386 M = 7.3726

Hence, number of moles of Iodine ion

= 3 x 7.3726 = 22.1178 milimol

Total milimol of iodide = 23.8144 + 22.1178 = 45.9322 milimol

Total volume = 24.4 + 19.1 = 43.5 mL

Hence,

Molarity = 45.9322 mM /43.5 mL = 1.056 M