Question

In the laboratory a student combines 28.7 mL of a 0.240 M barium chloride solution with 17.5 mL of a 0.331 M chromium(II) chloride solution. What is the final concentration of chloride anion ?

Answer 1

0.549 M

Step-by-step explanation:

Considering:

`Molarity=(Moles\ of\ solute)/(Volume\ of\ the\ solution)`

`Moles =Molarity * {Volume\ of\ the\ solution}`

Barium chloride will furnish chloride ions as:

`BaCl_2\rightarrow Ba^(2+)+2Cl^-`

Given :

For barium chloride :

Molarity = 0.240 M

Volume = 28.7 mL

The conversion of mL to L is shown below:

1 mL = 10⁻³ L

Thus, volume = 28.7×10⁻³ L

Thus, moles of chlorine furnished by barium chloride is twice the moles of barium chloride as shown below:

`Moles =2* 0.240 * {28.7* 10^(-3)}\ moles`

Moles of chloride ions by barium chloride = 0.013776 moles

Chromium(II) chloride will furnish chloride ions as:

`CrCl_2\rightarrow Cr^(2+)+2Cl^-`

Given :

For chromium(II) chloride :

Molarity = 0.331 M

Volume = 17.5 mL

The conversion of mL to L is shown below:

1 mL = 10⁻³ L

Thus, volume = 17.5×10⁻³ L

Thus, moles of chlorine furnished by chromium(II) chloride is twice the moles of chromium(II) chloride as shown below:

`Moles =2* 0.331 * {17.5* 10^(-3)}\ moles`

Moles of chloride ions by chromium(II) chloride = 0.011585 moles

Total moles = 0.013776 moles + 0.011585 moles = 0.025361 moles

Total volume = 28.7×10⁻³ L + 17.5×10⁻³ L = 46.2×10⁻³ L

Concentration of chloride ions is:

`Molarity=(Moles\ of\ solute)/(Volume\ of\ the\ solution)`

`Molarity_(Cl^-)=(0.025361)/(46.2* 10^(-3))`

The final concentration of chloride anion = 0.549 M