Chapter U4.89, Problem 4E

Interpretation Introduction

Interpretation:The missing values in the table are to be filled.

Concept introduction: The reagent in reaction that controls the amount of product formed is termed as limiting reagent. After completion of chemical reaction, the limiting reagent is fully utilized.

Answer to Problem 4E

The missing values in the table are filled as shown below:

  

Explanation of Solution

The reaction between $\text{NaOH}$ and $\text{HCl}$ is shown below:

  $\text{NaOH}+\text{HCl}\to \text{NaCl}+{\text{H}}_2\text{O}$

Here, the $\text{NaOH}$ and $\text{HCl}$ ratio is $1:1$ . In titration at equivalence point, the number of moles of acid is equal to the number of moles of base.

From the II data, the initial concentration of HCl is calculated as shown below:

  $\begin{array}{c}{n}_{\text{HCl}}=n_{\text{NaOH}}\\ M_{\text{HCl}}\cdot V_{\text{HCl}}=M_{\text{NaOH}}\cdot V_{\text{NaOH}}\\ M_{\text{HCl}}\cdot 100\text{mL}=0.10\text{M}\cdot 200\text{mL}\\ M_{\text{HCl}}\text{=}\frac{0.10\text{M}\cdot 200\text{mL}}{100\text{mL}}\\ =0.2\text{M}\end{array}$

From the III data, the number of moles of NaOH,

  $\begin{array}{c}{n}_{\text{NaOH}}=M_{\text{NaOH}}\cdot V_{\text{NaOH}}\\ =0.10\text{M}\cdot 200\text{mL}\left(\frac{1\text{L}}{1000\text{mL}}\right)\\ =0.02\text{mol}\end{array}$

The number of moles of HCl,

  $\begin{array}{c}{n}_{\text{HCl}}=n_{\text{NaOH}}\\ =0.02\text{mol}\end{array}$

The initial concentration of HCl,

  $\begin{array}{c}{n}_{\text{HCl}}=n_{\text{NaOH}}\\ M_{\text{HCl}}\cdot V_{\text{HCl}}=M_{\text{NaOH}}\cdot V_{\text{NaOH}}\\ M_{\text{HCl}}\cdot 50\text{mL}=0.10\text{M}\cdot 200\text{mL}\\ M_{\text{HCl}}\text{=}\frac{0.10\text{M}\cdot 200\text{mL}}{50\text{mL}}\\ =0.4\text{M}\end{array}$

From the IV data, the volume of NaOH,

  $\begin{array}{c}{n}_{\text{HCl}}=n_{\text{NaOH}}\\ M_{\text{HCl}}\cdot V_{\text{HCl}}=M_{\text{NaOH}}\cdot V_{\text{NaOH}}\\ 0.050\text{M}\cdot 50\text{mL}=0.10\text{M}\cdot V_{\text{NaOH}}\\ V_{\text{NaOH}}\text{=}\frac{0.050\text{M}\cdot 50\text{mL}}{0.10\text{M}}\\ =25\text{mL}\end{array}$

From the V data, the number of moles of NaOH,

  $\begin{array}{c}{n}_{\text{NaOH}}=M_{\text{NaOH}}\cdot V_{\text{NaOH}}\\ =0.10\text{M}\cdot 73\text{mL}\left(\frac{1\text{L}}{1000\text{mL}}\right)\\ =0.0073\text{mol}\end{array}$

The number of moles of HCl,

  $\begin{array}{c}{n}_{\text{HCl}}=n_{\text{NaOH}}\\ =0.0073\text{mol}\end{array}$

The initial concentration of HCl,

  $\begin{array}{c}{n}_{\text{HCl}}=n_{\text{NaOH}}\\ M_{\text{HCl}}\cdot V_{\text{HCl}}=M_{\text{NaOH}}\cdot V_{\text{NaOH}}\\ M_{\text{HCl}}\cdot 100\text{mL}=0.10\text{M}\cdot 73\text{mL}\\ M_{\text{HCl}}\text{=}\frac{0.10\text{M}\cdot 73\text{mL}}{100\text{mL}}\\ =0.073\text{M}\end{array}$

Therefore, the given is filled as shown below:

  

Conclusion

The missing values in the table are filled as shown below: