Question

A 83.0mL solution has 7.30g of MgCl2 and 4.20g of NaHCO3 dissolved within it.
(A.) What is the molarity of NaHCO3 in solution?
(B.) What is the molarity of chlorine ions in solution?
(C.) How many ions are there in solution?

Answer 1

C = n_substance/V_solution

n = m/M

M(NaHCO₃) = 23+1+12+16* 3 = 84 g/mole

n(NaHCO₃) = m/M = 4.2/84 = 0.05 mole

A) C(NaHCO₃) = n_substance/V_solution = 0.05/0.083 = 0.6 mole/l

0.077 mole x mole

MgCl₂ -> Mg²⁺ + 2Cl^-

1 mole 2 mole

0.077/1 = x/2

x = (0.077*2)/1 = 0.154 mole = n(Cl^-)

B) C(Cl^-) = n/V = 0.154/0.083 = 1.86 mole/l

C)MgCl₂ + 2NaHCO₃ -> MgCO₃ + 2NaCl + CO₂ + H₂O

Mg²⁺ + 2Cl^- + 2Na⁺ + 2HCO₃^- -> MgCO₃ + 2Na⁺ + 2Cl^- + CO₂ + H₂O

quantity of ions = 2 ions of Na⁺ + 2 ions of Cl^- = 4