We have to determine the concentration of iodide ions in a 0.193 M solution of barium iodide.
Barium iodide is made up of barium and iodide.
Barium is a group 2A metal which forms a cation with a charge of +2: Ba2+
Iodine is a group 7A nonmetal which forms an anion with a charge of -1: I-
To make a formula for the compound, we have to cross multiply the charges.
The formula is: BaI2
Barium iodide dissociates into its ions in a solution.
BaI2(aq) à Ba2+(aq) + 2 I-(aq)
The concentration of iodide ions in a 0.193 M solution of barium iodide is __________.
a. 0.193 M
b. 0.386 M
c. 0.0965 M
d. 0.579 M
e. 0.0643 M
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