Question

How many grams of SrO should be dissolved in sufficient water to make $2.00$L of a solution with a pH = $11.00$?

Answer 1

Hint:To answer this question first we will write the reaction of SrO with water. We will determine the pOH and by using pOH we will determine the hydroxide ion concentration. From the hydroxide ion concentration we will determine the concentration of strontium hydroxide. Then by using molarity formula we will determine the mole of hydroxide ions. By using mole of hydroxide ion we will determine the mole of strontium hydroxide and then mole of strontium oxide. By using mole formula, we can determine the mass of SrO.

Complete step-by-step answer:On dissolving SrO in sufficient water, the strontium hydroxide forms. The reaction is shown as follows:
${\text{SrO}}\,{\text{ + }}\,{{\text{H}}_{\text{2}}}{\text{O}}\,\, \to \,{\text{Sr(OH}}{{\text{)}}_{\text{2}}}$
Strontium hydroxide is an ionic compound. It dissociates in water as follows:
${\text{Sr(OH}}{{\text{)}}_{\text{2}}} \to \,{\text{S}}{{\text{r}}^{2 + }}\, + \,2{\text{O}}{{\text{H}}^ - }$
It produces hydroxide ions in water. So, we have to determine the concentration of hydroxide ions to determine the concentration of SrO finally.
The relation between pH and pOH is as follows:
${\text{pH}}\,{\text{ + }}\,{\text{pOH}}\,{\text{ = }}\,{\text{14}}$
On substituting $11.00$ for pH,
${\text{11}}\,{\text{ + }}\,{\text{pOH}}\,{\text{ = }}\,{\text{14}}$
$\Rightarrow {\text{pOH}}\,{\text{ = }}\,{\text{14}} - {\text{11}}$
$\Rightarrow {\text{pOH}}\,{\text{ = }}\,3$
The relation between pOH and hydroxide ion concentration is as follows:
${\text{pOH}}\,{\text{ = }}\, - \log [{\text{O}}{{\text{H}}^ - }]$
On substituting $3$ for pOH,
${\text{3}}\,{\text{ = }}\, - \log [{\text{O}}{{\text{H}}^ - }]$
$\Rightarrow [{\text{O}}{{\text{H}}^ - }]\, = \,\,{10^{ - {\text{3}}}}$
So, the concentration of hydroxide ion is ${10^{ - 3}}$ mol per litre.
The formula of molarity is as follows:
${\text{molarity}}\,{\text{ = }}\,\dfrac{{{\text{moles}}\,{\text{of}}\,{\text{solute}}}}{{{\text{volume}}\,{\text{of}}\,{\text{solution}}}}$
On substituting ${10^{ - 3}}$ mol per litre for molarity and $2.00$L for volume of the solution,
${10^{ - 3}}\,{\text{mol/L}}\,{\text{ = }}\,\dfrac{{{\text{moles}}\,{\text{of}}\,{\text{solute}}}}{{{\text{2}}{\text{.00}}\,{\text{L}}}}$
Moles of solute (hydroxide ion) $\, = \,{10^{ - 3}}\,{\text{mol/L}}\, \times \,{\text{2}}{\text{.00}}\,{\text{L}}$
Moles of solute (hydroxide ion) $ = \,\,{\text{2}} \times \,\,{10^{ - 3}}\,{\text{mol}}$
So, the moles of hydroxide ion in solution is ${\text{2}} \times \,\,{10^{ - 3}}$.
From the equation ${\text{Sr(OH}}{{\text{)}}_{\text{2}}} \to \,{\text{S}}{{\text{r}}^{2 + }}\, + \,2{\text{O}}{{\text{H}}^ - }$ , we can say that two moles of hydroxide ions, ${\text{O}}{{\text{H}}^ - }$ are produced by one mole of ${\text{Sr(OH}}{{\text{)}}_{\text{2}}}$ so, ${\text{2}} \times \,\,{10^{ - 3}}$ moles of hydroxide ions, ${\text{O}}{{\text{H}}^ - }$ will be produced by,
$2$ mole ${\text{O}}{{\text{H}}^ - }$ = $1$ mole ${\text{Sr(OH}}{{\text{)}}_{\text{2}}}$
${\text{2}} \times \,\,{10^{ - 3}}$mole ${\text{O}}{{\text{H}}^ - }$ = ${\text{1}} \times \,\,{10^{ - 3}}$ mole ${\text{Sr(OH}}{{\text{)}}_{\text{2}}}$
From the equation ${\text{SrO}}\,{\text{ + }}\,{{\text{H}}_{\text{2}}}{\text{O}}\,\, \to \,{\text{Sr(OH}}{{\text{)}}_{\text{2}}}$, we can say that one mole ${\text{Sr(OH}}{{\text{)}}_{\text{2}}}$ is produced by one mole of SrO so, ${\text{1}} \times \,\,{10^{ - 3}}$ mole of ${\text{Sr(OH}}{{\text{)}}_{\text{2}}}$ will be produced by,
$1$mole ${\text{Sr(OH}}{{\text{)}}_{\text{2}}}$ = $1$ mole SrO
${\text{1}} \times \,\,{10^{ - 3}}$mole ${\text{Sr(OH}}{{\text{)}}_{\text{2}}}$ = ${\text{1}} \times \,\,{10^{ - 3}}$ mole SrO
So, the moles of SrO in solution is ${\text{1}} \times \,\,{10^{ - 3}}$.
The relation between mole and mass is as follows:
${\text{mole}}\,{\text{ = }}\,\dfrac{{{\text{mass}}}}{{{\text{molar}}\,{\text{mass}}}}$
The molar mass of SrO is $64.06$ g/mol.
On substituting ${{1}} \times \,\,{10^{ - 3}}$ for moles of SrO and $64.06$ g/mol for molar mass of SrO,
${{1}} \times \,\,{10^{ - 3}}\,{\text{mol}}\,{\text{ = }}\,\dfrac{{{\text{mass}}}}{{{\text{64}}{\text{.06}}\,{\text{g/mol}}}}$
$\Rightarrow {\text{mass}}\, = \,{\text{1}} \times \,\,{10^{ - 3}}\,{\text{mol}}\,\,{\text{64}}{\text{.06}}\,{\text{g/mol}}$
$\therefore {\text{mass}}\, = \,0.0641\,{\text{g}}$
So, the mass of SrO is $0.0641$ gram.
Therefore, $0.0641$ grams of SrO should be dissolved in sufficient water.

Note:The molarity is defined as the moles of solute dissolved in per litre of the solution. The unit of molarity is mol/L. For the stoichiometric comparison a balanced chemical equation is necessary. Molar mass of the SrO is the sum of atomic mass of strontium atom and oxygen atom. As strontium hydroxide produces hydroxide ions in water, so it is a base.