Chapter 14, Problem 40E

Interpretation Introduction

Interpretation:

The volume of the barium hydroxide $\left(\text{Ba}{\left(\text{OH}\right)}_{\text{2}}\right)$ solution used to neutralize the potassium hydrogen phthalate $\left({\text{KHC}}_{\text{8}}{\text{H}}_4{\text{O}}_{\text{4}}\right)$ is to be stated.

Concept introduction:

Neutralization reactions are the reactions in which the acid or base is neutralizing another base and acid.

The unknown value of any property can be find out using the stoichiometry from the balanced chemical equation.

Answer to Problem 40E

The volume of $\text{Ba}{\left(\text{OH}\right)}_{\text{2}}$ solution used to neutralize ${\text{KHC}}_{\text{8}}{\text{H}}_4{\text{O}}_{\text{4}}$ is $27.0\text{ mL}$.

Explanation of Solution

Stoichiometry of the reaction comes in and plays a very important role in doing the calculation of measurable quantities. Stoichiometry is nothing but a comparison of quantities of one species with the other on the basis of a balanced chemical equation. The numerical value placed in from of the compounds tells about the number of moles of that compound reacting or producing in the reaction.

The balanced chemical equation of the reaction that happens between the barium hydroxide and potassium hydrogen phthalate is shown below.

${\text{2 KHC}}_{\text{8}}{\text{H}}_4{\text{O}}_{\text{4}}\left(aq\right)+\text{Ba}{\left(\text{OH}\right)}_{\text{2}}\left(aq\right)\to {\text{BaK}}_{\text{2}}{\left({\text{C}}_{\text{8}}{\text{H}}_4{\text{O}}_{\text{4}}\right)}_2\left(aq\right)+{\text{2 H}}_{\text{2}}\text{O}\left(l\right)$

Molarity is defined as the number of moles of solute per unit volume in liters of solution.

The formula of molarity in terms of volume in milliliters is given below.

It is multiplied by 1000 to convert the volume in milliliters to liters.

$\text{Molarity}=\frac{\text{Number of moles }\left(\text{n}\right)}{\text{Volume in milliliters (V)}}\times 1000$…(1)

The number of moles of any substance is related to its mass by the formula shown below.

$\text{Number of moles }\left(\text{n}\right)=\frac{\text{Mass of substance (m)}}{\text{Molar mass (M)}}$…(2)

Calculate the number of moles of ${\text{KHC}}_{\text{8}}{\text{H}}_4{\text{O}}_{\text{4}}$ using equation (2).

The molar mass of ${\text{KHC}}_{\text{8}}{\text{H}}_4{\text{O}}_{\text{4}}$ is given $204.23\text{g}/\text{mol}$.

The mass of ${\text{KHC}}_{\text{8}}{\text{H}}_4{\text{O}}_{\text{4}}$ neutralized is $1.655\text{ g}$.

Substitute the values of m and M in equation (2).

$\text{Number of moles }\left(\text{n}\right)=\frac{1.655\text{ g}}{204.23\text{g}/\text{mol}}$

The number of moles of barium hydroxide $\left(\text{Ba}{\left(\text{OH}\right)}_{\text{2}}\right)$ can be calculated using equation (1).

Let us suppose the volume of barium hydroxide $\left(\text{Ba}{\left(\text{OH}\right)}_{\text{2}}\right)$ is V(mL).

The molarity of the barium hydroxide $\left(\text{Ba}{\left(\text{OH}\right)}_{\text{2}}\right)$ solution used is $0.150M$.

Substitute the values of molarity and volume in equation (1).

$0.150M=\frac{\text{Number of moles }\left(\text{n}\right)}{\text{V}\left(\text{mL}\right)}\times 1000$

Rearrange to find out the number of moles as shown below.

$\text{Number of moles }\left(\text{n}\right)=\frac{\text{0}\text{.150 }M\times \text{V}\left(\text{mL}\right)}{1000}$

Compare the moles of both the $\text{Ba}{\left(\text{OH}\right)}_{\text{2}}$ and ${\text{KHC}}_{\text{8}}{\text{H}}_4{\text{O}}_{\text{4}}$ using stoichiometry relation.

One mole of $\text{Ba}{\left(\text{OH}\right)}_{\text{2}}$ is reacting with two moles of ${\text{KHC}}_{\text{8}}{\text{H}}_4{\text{O}}_{\text{4}}$.

Therefore, each one mole of $\text{Ba}{\left(\text{OH}\right)}_{\text{2}}$ is reacting with two moles of ${\text{KHC}}_{\text{8}}{\text{H}}_4{\text{O}}_{\text{4}}$.

$\text{2}\times \text{Number of moles}\left(\text{Ba}{\left(\text{OH}\right)}_{\text{2}}\right)=\text{Number of moles}\left({\text{KHC}}_{\text{8}}{\text{H}}_4{\text{O}}_{\text{4}}\right)$

Substitute the values of the number of moles of both species.

$\begin{array}{rll}\text{2}\times \text{Number of moles}\left(\text{Ba}{\left(\text{OH}\right)}_{\text{2}}\right)& =& \text{Number of moles}\left({\text{KHC}}_{\text{8}}{\text{H}}_4{\text{O}}_{\text{4}}\right)\\ 2\times \frac{\text{0}\text{.150 }M\times \text{V}\left(\text{mL}\right)}{1000}& =& \frac{1.655\text{ g}}{204.23\text{g}/\text{mol}}\end{array}$

Rearrange to calculate the volume as shown below.

$\begin{array}{rll}\text{V}\left(\text{mL}\right)& =& \frac{\left(1.655\text{ g}\right)\times 1000}{204.23\text{g}/\text{mol}\times \text{0}\text{.150 }M\times 2}\\ & =& 27.01\text{ mL}\\ & \approx & 27.0\text{ mL}\end{array}$

The volume of $\text{Ba}{\left(\text{OH}\right)}_{\text{2}}$ solution used to neutralize ${\text{KHC}}_{\text{8}}{\text{H}}_4{\text{O}}_{\text{4}}$ is $27.0\text{ mL}$.

Conclusion

The volume of $\text{Ba}{\left(\text{OH}\right)}_{\text{2}}$ solution used to neutralize ${\text{KHC}}_{\text{8}}{\text{H}}_4{\text{O}}_{\text{4}}$ is calculated as $27.0\text{ mL}$.