Question

A 23.33-mL solution containing 1.718 g Mg(NO₃)₂ is mixed with a 30.50-mL solution containing 1.467 g NaOH. Calculate the concentrations of the ions remaining in solution after the reaction is complete. Assume volumes are additive. If a species fully precipitates, type 0.

a.) ______ M Mg²⁺

b.) ______ M NO₃^-

c.) ______ M Na⁺

d.) ______ M OH^-

Answer 1

The correct answer is as shown below.

Mg(NO3)2 + 2 NaOH ? Mg(OH)2(s) + 2 NaNO3

(1.718 g Mg(NO3)2) / (148.3 g Mg(NO3)2/mol) = 0.0116 mol Mg(NO3)2
(1.467 g NaOH) / (39.99715 g NaOH/mol) = 0.0367 mol NaOH

0.0116 mole of Mg(NO3)2 would react completely with 0.0116 x (2/1) = 0.0232 mole of NaOH, but there is more NaOH present than that, so NaOH is in excess and Mg(NO3)2 is the limiting reactant.

(Neglecting the Ksp of Mg(OH)2) sInce all of the Mg is bound in the insoluble Mg(OH)2 at the end of the reaction, there are no Mg{2+} ions present then.

(0.0367 mol NaOH) x (1 mol Na{+} / 1 mol NaOH) / (0.02333 L + 0.03050 L) = 0.6818 mol/L Na{+}

((0.0367 mol NaOH initially) - ( 0.0232 mol NaOH reacted)) x (1 mol OH{-} / 1 mol NaOH) /
(0.02333 L + 0.0305 L) = 0.251 mol/L OH{-}

(0.0116 mol Mg(NO3)2) x (2 mol NO3{-} / 1 mol Mg(NO3)2) / (0.02333 L + 0.0305 L) =
0.215 mol/L NO3{-}

Thus, the final concentration of ions in solution would be,

a.) ___0.0_ M Mg²⁺

b.) _0.215 _ M NO₃^-

c.) _0.6818 _ M Na⁺

d.) _0.251 _ M OH^-