0.522 g of solid Mg(OH)₂ are dissolved in 25.0mL of a 4.0 M HNO₃ solution. At the end, the solution remains acidic.

1) Write the chemical equation of this reaction

2) If you assume the reaction is complete, calculate the concentration of Mg²⁺, NO₃^- and H⁺.

Question

0.522 g of solid Mg(OH)₂ are dissolved in 25.0mL of a 4.0 M HNO₃ solution. At the end, the solution remains acidic.

1) Write the chemical equation of this reaction

2) If you assume the reaction is complete, calculate the concentration of Mg²⁺, NO₃^- and H⁺.

| General Chemistry 2

Accepted Answer

1) Mg(OH)₂ + 2 HNO₃ → 2 H₂O + Mg(NO₃)₂

2) The solution remains acidic: Mg(OH)₂ is the limiting reactant

Concentration of Mg²⁺:

According to the equation, if the reaction is complete:

n_Mg(NO3)2 = n_Mg(OH)2 = $\frac{m_{Mg (OH) 2}}{M_{Mg (OH) 2}}$

n_Mg(NO3)2 = $\frac{0.612}{58.309}$ = 8.95 x 10^-3 mol

Mg(NO₃)₂ = Mg²⁺ + 2 NO₃^-

[Mg²⁺] = $\frac{n_{Mg 2 +}}{V_{sol}}$ = $\frac{n_{Mg (NO 3) 2}}{V_{sol}}$ = $\frac{8.95 \times 10^{- 3}}{25.0 \times 10^{- 3}}$ = 3.58 x 10^-1 M

Concentration of NO₃^-:

According to the equation 2 moles of NO₃^- are consumed to form 2 moles of NO₃^-
⇒ [NO₃^-] is unchanged ⇒ [NO₃^-] = [HNO₃]₀ = 4.0 M

Concentration of H⁺:

According to the equation: 2 moles of HNO₃ react with 1 mole of Mg(OH)₂.

Mg(OH)₂ is the limiting reactant:
n_HNO3 = n_HNO3 (t=0) – 2 x n_Mg(OH)2 (t=0)

n_HNO3 = [HNO₃] x V_sol – 2 x $\frac{m_{Mg (OH) 2}}{M_{Mg (OH) 2}}$

n_HNO3 = 4.0 x 25.0 x 10^-3 – 2 x $\frac{0.522}{58.309}$ = 8.21 x 10^-2 mol

HNO₃ = H⁺ + NO₃^-

[H⁺] = $\frac{n_{H +}}{V_{sol}}$ = $\frac{n_{HNO 3}}{V_{sol}}$ = $\frac{5.83 \times 10^{- 2}}{25.0 \times 10^{- 3}}$ = 3.28 M

Midterm 2 - Question 2 | General Chemistry 2