Answer to Question #164972 in Chemistry for James

Question

in a reaction 35.0 mL of 0.718 M copper II chloride reacts with 45.0 mL of 0.429 M potassium phosphate. Determine the molarity of each of the spectator ions.

Answer 1

Copper II chloride (CaCl₂) solution (35.0 mL) contains 35.0 mL × 0.718 M = 0.035 L × 0.718 mol/L = 0.025 mol.

From here, the amounts of ions in the initial solution are:

n(Ca²⁺) = 0.025 mol

n(Cl^-) = 0.025 × 2 = 0.05 mol

Potassium phosphate (K₃PO₄) solution (45.0 mL) contains 45.0 mL × 0.429 M = 0.045 L × 0.429 mol/L = 0.019 mol.

From here, the amounts of ions in the initial solution are:

n(K⁺) = 0.019 mol × 3 = 0.057 mol

n(PO₄^3-) = 0.019 mol

The final volume of the solution is:

V = 35.0 mL + 45.0 mL = 80.0 mL = 0.080 L

From here, the molarity of each ion in the final solution is:

c(Ca²⁺) = n(Ca²⁺) / V = 0.025 mol / 0.080 L = 0.31 M

c(Cl^-) = n(Cl^-) / V = 0.05 mol / 0.080 L = 0.625 M

c(K⁺) = n(K⁺) / V = 0.057 mol / 0.080 L = 0.071 M

c(PO₄^3-) = n(PO₄^3-) / V = 0.019 mol / 0.080 L = 0.24 M