Chapter 23, Problem 4STP

Interpretation Introduction

Interpretation:

The mass of NaOH in grams required to completely neutralize 50.0 mL of 0.100 M HCl needs to be determined.

Concept introduction:

The reaction between HCl and NaOH can be given as,

$\text{NaOH}\left(\text{aq}\right)\text{+}\text{HCl}\left(\text{aq}\right)\to \text{NaCl}\left(\text{aq}\right)\text{+}{\text{H}}_{\text{2}}\text{O}\left(\text{l}\right)$

NaOH reacts with HCl to give NaCl and ${\text{H}}_{\text{2}}\text{O}$ as products.

Answer to Problem 4STP

0.200 g of NaOH is required to completely neutralize 50.0 mL of 0.100 M HCl.

Explanation of Solution

The reaction between HCl and NaOH can be given as,

$\text{NaOH}\left(\text{aq}\right)\text{+}\text{HCl}\left(\text{aq}\right)\to \text{NaCl}\left(\text{aq}\right)\text{+}{\text{H}}_{\text{2}}\text{O}\left(\text{l}\right)$

So, we can calculate the amount of NaOH in grams required to completely neutralize 50.0 mL of 0.100 M HCl by using the following method.

$\begin{array}{c}\text{Mass}\text{of}\text{NaOH=}\text{50}\text{.0}\text{mL}\text{HCl×}\frac{\text{1}\text{L}}{\text{1000}\text{mL}}\text{×}\frac{\text{0}\text{.100}\text{mol}\text{HCl}}{\text{1}\text{L}}\text{×}\frac{\text{1}\text{mol}\text{NaOH}}{\text{1}\text{mol}\text{HCl}}\text{×}\frac{\text{40}\text{.0}\text{g}\text{NaOH}}{\text{1}\text{mol}\text{of}\text{NaOH}}\\ \text{=}\text{0}\text{.200}\text{g}\text{NaOH}\end{array}$

Conclusion

The amount of NaOH in grams required to completely neutralize 50.0 mL of 0.100 M HCl is 0.200 g.