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

Answer : The final concentration of chloride ion is, 0.549 M

Explanation :

$Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}$

$Moles =Molarity \times {Volume\ of\ the\ solution}$

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:

$Moles =2\times 0.240 \times {28.7\times 10^{-3}}\ moles$

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:

$Moles =2\times 0.331 \times {17.5\times 10^{-3}}\ moles$

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:

$Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}$

$\text{Molarity of chloride anion}=\frac{0.025361}{46.2\times 10^{-3}}=0.549M$

Therefore, the final concentration of chloride anion is 0.549 M