Question

A 30.omL sample an unknown strong base is neutralized after the addition of 12.0mL of a 0.150 ${\mathbf{MHNO}}_{\mathbf{3}}$solution. If the unknown base concentration is 0.0300M, give some possible identities for the unknown base.

Short answer

The possible identities are$\mathrm{Ba}{\left(\mathrm{OH}\right)}_2,\mathrm{Ca}{\left(\mathrm{OH}\right)}_2,\text{and}\mathrm{Mg}{\left(\mathrm{OH}\right)}_2$ .

Step-by-step solution

Given data

Volume of ${\mathrm{HNO}}_3$ solution=12.0mL

Concentration of ${\mathrm{HNO}}_3$solution =0.150M

Calculation of moles

No. of moles of ${\mathrm{HNO}}_3$ present in 12.0mL of 0.150M ${\mathrm{HNO}}_3$ solution

$\begin{array}{l}=12.0\mathrm{mL}\times \frac{1\mathrm{L}}{100\mathrm{mL}}\times 0.150\mathrm{M}\\ =12.0\mathrm{mL}\times \frac{1\mathrm{L}}{1000\mathrm{mL}}\times 0.150\frac{\mathrm{mol}}{\mathrm{L}}\\ =1.80\times {10}^{-3}\mathrm{mol}\end{array}$

Calculation of ratio of moles

Volume of unknown base solution=30.0mL

Concentration of unknown base solution =0.0300M

No. of moles of unknown base present in 30.0mL of 0.0300M solution

$\begin{array}{l}=30.0\mathrm{mL}\times \frac{1\mathrm{L}}{100\mathrm{mL}}\times 0.300\mathrm{M}\\ =30.0\mathrm{mL}\times \frac{1\mathrm{L}}{1000\mathrm{mL}}\times 0.300\frac{\mathrm{mol}}{\mathrm{L}}\\ =9.00\times {10}^{-4}\mathrm{mol}\end{array}$

Ratio of no. of molesof acid to the no. of moles of base:

$\begin{array}{c}\frac{{\mathrm{n}}_{\mathrm{acid}}}{{\mathrm{n}}_{\mathrm{base}}}=\frac{1.80\times {10}^{-3}\text{moleacid}}{9.00\times {10}^{-4}\text{molebase}}\\ =2\end{array}$

Possible identities

The ratio of the moles of acid to the moles of base shows that there are two times as many moles of acid as there are to the base. This means that our base must provide two hydroxyl ions for each molecule of acid. Therefore, our base could be any compound that provides two molecules of hydroxyl ions.

Hence, some of the possible identities for the base are

$\mathrm{Ba}{\left(\mathrm{OH}\right)}_2,\mathrm{Ca}{\left(\mathrm{OH}\right)}_2,\text{and}\mathrm{Mg}{\left(\mathrm{OH}\right)}_2$.