Question

In: Chemistry

In the laboratory a student combines 49.7 mL of a 0.133 M magnesium iodide solution with...

In the laboratory a student combines 49.7 mL of a 0.133 M magnesium iodide solution with 28.0 mL of a 0.562 M nickel(II) iodide solution.  

What is the final concentration of iodide anion ?

___ M

Solutions

Expert Solution

final concentration of iodide anion = 0.575 M

Explanation

magnesium iodide MgI2

nickel (II) iodide NiI2

concentration of MgI2 = 0.133 M

volume MgI2 = 49.7 mL

moles MgI2 = (concentration of MgI2) * (volume MgI2)

moles MgI2 = (0.133 M) * (49.7 mL)

moles MgI2 = 6.6101 mmol

moles I- from MgI2 = 2 * (moles MgI2)

moles I- from MgI2 = 2 * (6.6101 mmol)

moles I- from MgI2 = 13.2202 mmol

concentration of NiI2 = 0.562 M

volume NiI2 = 28.0 mL

moles NiI2 = (concentration of NiI2) * (volume NiI2)

moles NiI2 = (0.562 M) * (28.0 mL)

moles NiI2 = 15.736 mmol

moles I- from NiI2 = 2 * (moles NiI2)

moles I- from NiI2 = 2 * (15.736 mmol)

moles I- from NiI2 = 31.472 mmol

Total moles I- = (moles I- from MgI2) + (moles I- from NiI2)

Total moles I- = (13.2202 mmol) + (31.472 mmol)

Total moles I- = 44.6922 mmol

Total volume = (volume MgI2) + (volume NiI2)

Total volume = (49.7 mL) + (28.0 mL)

Total volume = 77.7 mL

Final concentration = (Total moles I-) / (Total volume)

Final concentration = (44.6922 mmol) / (77.7 mL)

Final concentration = 0.575 M


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