Question

A chemist obtains 500.0 mL of a solution containing an unknown concentration of calcium iodide, CaI2. He pipets 15 mL of this solution into a 100 mL volumetric flask and dilutes to the mark. He then pipettes 10 mL of this diluted solution into a 25 mL volumetric flask and dilutes to the mark. He analyzes some of the solution from the final volumetric flask and finds that the iodide ion concentration is 0.41 M. (Note: in solution, calcium iodide breaks apart into one Ca2+ ion for every two I- ions, so a solution that is 1.0 M in CaI2 is 2.0 M in I-.) Determine the molar concentration of calcium iodide in the original solution. LaTeX: CaI _{2} (s)→ Ca^{ 2+}(aq)+2 I^{-}(aq)

Answer 1

3.41735M

Step-by-step explanation:

Step1: Balance the equation

CaI2 (s) → Ca2+ (aq ) + 2I- (aq)

We have 15 mL of the original sample and pipet it in a 100mL solution.

100/15 = 6.67 ⇒ this is a 1:6.67 dilution

We take 10mL of this dilution into a 25 mL flask ⇒this is a 1:2.5 dilution

6.67 * 2.5 = 16.675

⇒the total dilution is 1:16.675

Let's imagine the conc of CaI2 [CaI2] = 1M

⇒ 15mL of this 1M conc in a 100mL = 0.15M

⇒10mL of this 0.15 M in 25mL = 0.06M

1M original conc/ O.O6M final concentration = 16.67

⇒this means the dilution is 16.67 (1:16.67)

After analyzes we found that the iodide ion concentration = 0.41 M

⇒original conc / 0.41 M = 16.67

original conc I- = 16.67 * 0.41M = 6.8347 M

Since we have one Ca2+ ion for every 2 I- ions, we have to divide this concentration by 2 to find the [Ca2+]

6.8347 / 2 = 3.41735 M =[Ca2+] = [CaI2]

The concentration calcium iodide = 3.41735M