Determine the molar concentration of Ca^2+ and Cl- in a .075 M CaCl2 solution
To determine the molar concentration of Ca^2+ and Cl- in a 0.075 M CaCl2 solution, we need to consider the dissociation of CaCl2.
CaCl2 dissociates into Ca^2+ and 2 Cl- ions. This means that for every 1 mole of CaCl2 that dissolves, we get 1 mole of Ca^2+ ions and 2 moles of Cl- ions.
Therefore, the molar concentration of Ca^2+ in the CaCl2 solution is also 0.075 M.
On the other hand, since each CaCl2 molecule produces 2 Cl- ions, the molar concentration of Cl- is two times the molar concentration of CaCl2.
Molar concentration of Cl- = 2 * 0.075 M = 0.15 M
So, in a 0.075 M CaCl2 solution, the molar concentration of Ca^2+ and Cl- ions are both 0.075 M and 0.15 M, respectively.
To determine the molar concentration of Ca^2+ and Cl- in a CaCl2 solution, we need to consider the stoichiometry of the compound.
CaCl2 dissociates into one Ca^2+ ion and two Cl- ions in water. This means that for every CaCl2 molecule, we have one Ca^2+ ion and two Cl- ions.
Since the concentration of the CaCl2 solution is given as 0.075 M, it means that the concentration of Ca^2+ ions will also be 0.075 M because there is a 1:1 stoichiometric ratio between CaCl2 and Ca^2+.
However, because there are two Cl- ions for every Ca^2+ ion, the concentration of Cl- ions will be twice that of Ca^2+. Therefore, the concentration of Cl- ions in the CaCl2 solution will be 2 * 0.075 M = 0.15 M.
So, the molar concentration of Ca^2+ in the CaCl2 solution is 0.075 M, and the molar concentration of Cl- ions is 0.15 M.