In: Chemistry
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
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