Chapter 9, Problem 9.127QP

(a)

Interpretation Introduction

Interpretation:

The preparation of Magnesium hydroxide should be described and it solubility should be predicted and the volume of $\text{HCl}$ solution needed to react with given $\text{10 mL}$ of milk of magnesia should be calculated.

Concept introduction:

Precipitation reaction:

The precipitate is formed, when the two solutions mixed together is called precipitation reaction.

The mass of the precipitate is depends on the reactant masses.

Molarity:

The concentration of the solutions is given by the term of molarity and it is given by ratio between numbers of moles of solute present in litter of solution.

$\text{Molarity}\text{=}\frac{\text{No}\text{.mole}}{\text{volume}\text{(L)}}$

Mole:

The mole of compound is given by the ratio between taken mass of the compound and molar mass of the compound.

$\text{Mole}\text{=}\frac{\text{Massof}\text{the}\text{compound}}{\text{Molar}\text{mass}\text{of}\text{the}\text{compound}}$

Mole:

The mole of the solute is calculated by multiplication of concentration of solution and volume of the solution and it is,

$\text{Mole}\text{=}\text{Concentration}\text{(M)×}\text{volume}\text{(L)}$

Answer to Problem 9.127QP

Magnesium hydroxide is insoluble in water so this is prepared by the reaction of aqueous solution of ${\text{MgCl}}_{\text{2}}\text{ or Mg}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}$ with hydroxide ion containing solution like $\text{NaOH}$ to produce the precipitate of Magnesium hydroxide.

The net ionic equation for the preparation of Magnesium hydroxide is,

${\text{Mg}}^{\text{2+}}{}_{\text{(aq)}}{\text{+2OH}}^{\text{-}}{}_{\text{(aq)}}\to {\text{Mg(OH)}}_{\text{2}}{}_{\text{(s)}}$

Explanation of Solution

Describe the preparation of Magnesium hydroxide and predict and it solubility in water

1. (a) Magnesium hydroxide is insoluble in water so this is prepared by the reaction of aqueous solution of ${\text{MgCl}}_{\text{2}}\text{ or Mg}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}$ with hydroxide ion containing solution like $\text{NaOH}$ to produce the precipitate of Magnesium hydroxide.

• The net ionic equation for the preparation of Magnesium hydroxide is,

• ${\text{Mg}}^{\text{2+}}{}_{\text{(aq)}}{\text{+2OH}}^{\text{-}}{}_{\text{(aq)}}\to {\text{Mg(OH)}}_{\text{2}}{}_{\text{(s)}}$

(b)

Interpretation Introduction

Interpretation:

The preparation of Magnesium hydroxide should be described and it solubility should be predicted and the volume of $\text{HCl}$ solution needed to react with given $\text{10 mL}$ of milk of magnesia should be calculated.

Concept introduction:

Precipitation reaction:

The precipitate is formed, when the two solutions mixed together is called precipitation reaction.

The mass of the precipitate is depends on the reactant masses.

Molarity:

The concentration of the solutions is given by the term of molarity and it is given by ratio between numbers of moles of solute present in litter of solution.

$\text{Molarity}\text{=}\frac{\text{No}\text{.mole}}{\text{volume}\text{(L)}}$

Mole:

The mole of compound is given by the ratio between taken mass of the compound and molar mass of the compound.

$\text{Mole}\text{=}\frac{\text{Massof}\text{the}\text{compound}}{\text{Molar}\text{mass}\text{of}\text{the}\text{compound}}$

Mole:

The mole of the solute is calculated by multiplication of concentration of solution and volume of the solution and it is,

$\text{Mole}\text{=}\text{Concentration}\text{(M)×}\text{volume}\text{(L)}$

Answer to Problem 9.127QP

The volume of $\text{HCl}$ solution needed to react with given $\text{10 mL}$ of milk of magnesia solution is $\text{0}\text{.78 L}$

Explanation of Solution

Record the given data,

Concentration of $\text{HCl}$ solution = $\text{0}\text{.035 M}$

Volume of ${\text{Mg(OH)}}_{\text{2}}$ solution = $\text{10}\text{.0 mL}$

Concentration of ${\text{Mg(OH)}}_{\text{2}}$ solution = $\text{0}{\text{.080 g Mg(OH)}}_{\text{2}}\text{/mL}$

The given volume and Concentrations are recorded as shown above.

Calculate the mole of ${\text{Mg(OH)}}_{\text{2}}$ in given milk of magnesia.

Molar mass of ${\text{Mg(OH)}}_{\text{2}}$ is $\text{58}\text{.326 g}$

The balance equation is,

${\text{Mg(OH)}}_{\text{2}}{\text{+2HCl}}_{\text{(aq)}}\to {\text{MgCl}}_{\text{2(aq)}}{\text{+H}}_{\text{2}}{\text{O}}_{\text{(l)}}$

$\begin{array}{l}{\text{Mg(OH)}}_{\text{2}}\text{mole}\text{=}\text{10}\text{.0}\text{mL×}\frac{\text{0}\text{.080}\text{g}{\text{Mg(OH)}}_{\text{2}}}{\text{1}\text{mol}\text{soln}}\text{×}\frac{\text{1}\text{mol}{\text{Mg(OH)}}_{\text{2}}}{\text{58}\text{.}\text{326}\text{g}{\text{Mg(OH)}}_{\text{2}}}\\ \text{=}\text{0}\text{.0137}\text{mole}\end{array}$

• The molar mass of ${\text{Mg(OH)}}_{\text{2}}$ and concentration of milk of magnesia are plugged in the above equation to give the mole of ${\text{Mg(OH)}}_{\text{2}}$ present in the given volume of milk of magnesia.
• The mole of ${\text{Mg(OH)}}_{\text{2}}$ present in the given volume of milk of magnesia is $\text{0}\text{.0137}\text{mole}$.

Determine the volume of $\text{HCl}$ solution needed to react with given $\text{10 mL}$of milk of magnesia.

$\begin{array}{l}\text{Moles}\text{of}\text{HCl}\text{reacted}\text{=}\text{0}\text{.0137}\text{mol}{\text{Mg(OH)}}_{\text{2}}\text{×}\frac{\text{2}\text{mol}\text{HCl}}{\text{1}\text{mol}{\text{Mg(OH)}}_{\text{2}}}\\ \text{=}\text{0}\text{.0274}\text{mole}\\ \\ \text{Volume}\text{of}\text{HCl}\text{=}\frac{\text{solute}\text{mole}}{\text{M}}\\ \text{=}\frac{\text{0}\text{.0274}\text{mol}\text{HCl}}{\text{0}\text{.035}\text{mol/L}}\\ =0.78\mathrm{L}\end{array}$

• Two moles of $\text{HCl}$ required to react with one mole of so required mole of solution is 1:2.
• The calculated mole of ${\text{Mg(OH)}}_{\text{2}}$ solution is plugged in the equation to give the required mole of $\text{HCl}$ solution to react ${\text{Mg(OH)}}_{\text{2}}$ with solution.
• The calculated $\text{HCl}$ mole of solution is divided by given molarity of $\text{HCl}$ solution to give volume of $\text{HCl}$ solution needed to react with given $\text{10 mL}$ of milk of magnesia
• The volume of $\text{HCl}$ solution needed to react with given $\text{10 mL}$ of milk of magnesia is $\text{0}\text{.78 L}$.