Question

(2.150x10^3)mL of dilute calcium chloride solution is prepared from (4.339x10^2)mL of (2.11x10^0)M solution. What is the final concentration of chloride ions

Answer 1

The final concentration of chloride ions in the dilute calcium chloride solution is approximately 8.526 x 10^0 M.

Given:

Initial concentration (M1) = 2.11 x 10^0 M

Initial volume (V1) = 4.339 x 10^2 mL

Final volume (V2) = 2.15 x 10^3 mL

We can use the dilution formula:

M1 * V1 = M2 * V2

Substitute the given values:

(2.11 x 10^0 M) * (4.339 x 10^2 mL) = M2 * (2.15 x 10^3 mL)

Now, solve for M2:

M2 = (2.11 x 10^0 M * 4.339 x 10^2 mL) / (2.15 x 10^3 mL)

Calculate the result:

M2 = (9.1619 / 2.15) x 10^0 M

M2 ≈ 4.263 x 10^0 M

Since calcium chloride dissociates into calcium ions and chloride ions in a 1:2 ratio, the final concentration of chloride ions (Cl-) is twice the calculated concentration of calcium chloride:

Final concentration of Cl- = 2 x M2

Final concentration of Cl- = 2 x (4.263 x 10^0 M)

Final concentration of Cl- ≈ 8.526 x 10^0 M