Question

Calculate the concentration of all ions present in the solution of strong electrolytes 0.15 m calcium chloride

Answer 1

The concentration of calcium ions (Ca2+) in a 0.15 M solution of calcium chloride is 0.15 M, and the concentration of chloride ions (Cl-) is 0.30 M.

Step-by-step explanation:

The concentration of ions in a solution of strong electrolytes like calcium chloride can be calculated by considering the stoichiometry of the dissolution reaction. Calcium chloride, CaCl2, dissociates completely into calcium ions, Ca2+, and chloride ions, Cl-, in water. Given an initial concentration of 0.15 M for CaCl2, for every mole of CaCl2 that dissolves, one mole of Ca2+ ions and two moles of Cl- ions are produced. Therefore, the concentration of Ca2+ is 0.15 M, and the concentration of Cl- is 0.15 M × 2, or 0.30 M.

Answer 2

Calcium chloride, Ca Cl2 is a strong electrolyte which means that if fully ionizes. This is the ionization equation:

CaCl2 ---> Ca(2+) + 2 Cl(-).

So, per each mole of CaCl2 1 mole of Ca(2+) and 2 moles of Cl(-) are produced.

That means that the concentration of Ca(2+) is the same concentration of CaCl2 and the concentration of Cl(-) is twice the concentration of CaCl2.

=> [Ca(2+)] = 0.15 M, and [Cl-] = 2 * 0.15M = 0.30M

Answer:

[Ca(2+)] = 0.15M

[Cl(-)] = 0.30 M