What is the concentration of chloride ion made from an $80 \cdot m L$ $80mL$ volume of $0.125 \cdot m o l \cdot L^{- 1}$ $0.125molL^-1$ $C a C l_{2} (a q)$ $CaCl_2(aq)$ that is diluted to a $100 \cdot m L$ $100mL$ volume?

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Answer 1 · 2016-01-26T16:24:30

$[C l^{-}]$ $[Cl^-]$ $=$ $=$ $\frac{Moles of chloride ion}{Volume of solution}$ $("Moles of chloride ion")/("Volume of solution")$

$\frac{Moles of chloride}{Volume of solution}$ $"Moles of chloride"/"Volume of solution"$

We note that in the calcium chloride solution, each equiv calcium chloride contributes 2 equiv halide ion:

$[C l^{-}]$ $[Cl^-]$ $=$ $=$ $(2xx0.125*mol*cancelL^-1xx80xx10^-3cancelL)/(100xx10^-3L)$

$=$ $??mol*L^-1$

What is the concentration of chloride ion made from an 80*mL80⋅mL80*mL volume of 0.125*mol*L^-10.125⋅mol⋅L−10.125*mol*L^-1 CaCl_2(aq)CaCl2(aq)CaCl_2(aq) that is diluted to a 100*mL100⋅mL100*mL volume?