Question

Indicate the concentration of each ion present in the solution formed by mixing

c) 3.60g of KCL in 75.0mL of .250M CaCl2 solution.

Answer 1

mass of KCl = 3.60 g

molar mass of KCl = 74.55 g/mol

moles of KCl = 3.60 / 74.55

                      = 0.0483

moles of K = 0.0483

concentration of [K+] = moles / total volume

                                  = 0.0483 / 0.075

                                   = 0.644 M

concentration of [K+] = 0.644 M

volume of CaCl2 = 75.0 mL = 0.075 L

molarity = 0.250 M

moles of CaCl2 = molarity x volume

                          = 0.250 x 0.075

                          = 0.01875

moles of Ca = 0.01875

concentration of [Ca+2] = 0.01875 / 0.075

                                       = 0.25 M

concentration of [Ca+2] = 0.25 M

moles of Cl- = Cl- from KCl + 2 Cl- from CaCl2

                    = 0.0483 + 2 x 0.01875

                    = 0.0858

concentration of [Cl-] = 0.0858 / 0.075

                                  = 1.144 M

concentration of [Cl-] = 1.144 M