Chapter 14, Problem 5ST

Interpretation Introduction

Interpretation:

The volume of barium hydroxide required to neutralize $0.100M\text{HCl}$ is to be identified.

Concept introduction:

Molarity is defined as the number of moles present in one liter of solution. A general expression is shown below.

$\text{Molarity}=\frac{\text{Moles}\text{of}\text{solute}}{\text{Volume}\text{of}\text{solution}}$

Measuring unit of molarity is $\text{mol}/\text{L}$ or $\text{molar}\left(\text{M}\right)$.

Answer to Problem 5ST

Correct answer:

The correct option is (a).

Explanation of Solution

Reason for correct answer:

The balanced neutralization reaction between $\text{HCl}$ and $\text{Ba}{\left(\text{OH}\right)}_{\text{2}}$ is shown below.

$\text{2HCl}\left(aq\right)+\text{Ba}{\left(\text{OH}\right)}_{\text{2}}\left(aq\right)\to {\text{BaCl}}_{\text{2}}\left(aq\right)+2{\text{H}}_{\text{2}}\text{O}\left(l\right)$

The number of moles of $\text{HCl}$ in terms of molarity is calculated by the formula shown below.

$n_{\text{HCl}}=M_{\text{HCl}}\times V_{\text{HCl}}$ …(1)

Where,

• $n_{\text{HCl}}$ is the number of moles of $\text{HCl}$.

• $M_{\text{HCl}}$ is the molarity of $\text{HCl}$.

• $V_{\text{HCl}}$ is the volume of $\text{HCl}$ required in the neutralization reaction.

The value of $M_{\text{HCl}}$ is $\text{0}\text{.100}\text{M}$ and the value of $V_{\text{HCl}}$ is $\text{25}\text{mL}$. Substitute the values of the corresponding variable in equation (1) as shown below.

$\begin{array}{rll}{n}_{\text{HCl}}& =& M_{\text{HCl}}\times V_{\text{HCl}}\\ & =& \text{0}\text{.100}\text{M}\times \left(\text{25}\text{mL}\times \frac{1\text{L}}{1000\text{mL}}\right)\\ & =& \text{0}\text{.100}\text{M}\times 0.025\text{L}\\ & =& 0.0025\end{array}$

So, the number of mole of $\text{HCl}$ is $0.0025$. In the balanced chemical reaction $2$ mol of $\text{HCl}$ neutralizes $1$ mol of $\text{Ba}{\left(\text{OH}\right)}_{\text{2}}$. So, the number of mole of $\text{Ba}{\left(\text{OH}\right)}_{\text{2}}$ will be half of the moles of $\text{HCl}$. The number of mole of $\text{Ba}{\left(\text{OH}\right)}_{\text{2}}$ is calculated as shown below.

$n_{\text{Ba}{\left(\text{OH}\right)}_{\text{2}}}=\frac{n_{\text{HCl}}}{2}$ …(2)

Substitute the value of $n_{\text{HCl}}$ in equation (2) as shown below.

$\begin{array}{rll}{n}_{\text{Ba}{\left(\text{OH}\right)}_{\text{2}}}& =& \frac{n_{\text{HCl}}}{2}\\ & =& \frac{0.0025}{2}\\ & =& 0.00125\end{array}$

The volume of the $\text{Ba}{\left(\text{OH}\right)}_{\text{2}}$ is calculated by the formula shown below.

$V_{\text{Ba}{\left(\text{OH}\right)}_{\text{2}}}=\frac{n_{\text{Ba}{\left(\text{OH}\right)}_{\text{2}}}}{M_{\text{Ba}{\left(\text{OH}\right)}_{\text{2}}}}$ …(3)

Where,

• $n_{\text{Ba}{\left(\text{OH}\right)}_{\text{2}}}$

is the number of moles of $\text{Ba}{\left(\text{OH}\right)}_{\text{2}}$.

• $M_{\text{Ba}{\left(\text{OH}\right)}_{\text{2}}}$ is the molarity of $\text{Ba}{\left(\text{OH}\right)}_{\text{2}}$.

• $V_{\text{Ba}{\left(\text{OH}\right)}_{\text{2}}}$ is the volume of $\text{Ba}{\left(\text{OH}\right)}_{\text{2}}$ required in the neutralization reaction.

The value of $n_{\text{Ba}{\left(\text{OH}\right)}_{\text{2}}}$ is $\text{0}\text{.00125}\text{mol}$ and the value of $M_{\text{Ba}{\left(\text{OH}\right)}_{\text{2}}}$ is $0.150M$. Substitute the values of the corresponding variable in equation (3) as shown below.

$\begin{array}{rll}{V}_{\text{Ba}{\left(\text{OH}\right)}_{\text{2}}}& =& \frac{\text{0}\text{.00125}\text{mol}}{0.150M}\\ & =& \left(8.33\times {10}^{-3}\text{L}\times \frac{1\text{mL}}{{10}^{-3}\text{L}}\right)\\ & =& 8.33\text{mL}\end{array}$

The volume of barium hydroxide required to neutralize $0.100M\text{HCl}$ is $8.33\text{mL}$.

The correct option is (a).

Reason for incorrect options:

(b) The volume of barium hydroxide required to neutralize $0.100M\text{HCl}$ is $8.33\text{mL}$.

Therefore, the option (b) is incorrect.

(c) The volume of barium hydroxide required to neutralize $0.100M\text{HCl}$ is $8.33\text{mL}$.

Therefore, the option (c) is incorrect.

(d) The volume of barium hydroxide required to neutralize $0.100M\text{HCl}$ is $8.33\text{mL}$.

Therefore, the option (d) is incorrect.

(e) The volume of barium hydroxide required to neutralize $0.100M\text{HCl}$ is $8.33\text{mL}$.

Therefore, the option (e) is incorrect.

Conclusion

The volume of barium hydroxide required to neutralize $0.100M\text{HCl}$ is $8.33\text{mL}$. Therefore, the correct option is (a).