Question

What volume (mL) of 0.135 M NaOH is required to neutralize 13.7 mL of 0.129 M HCl? a: 0.24 b: 13.1 c: 0.076 d: 6.55 e: 14.3

Answer 1

Answer:

The volume (mL) of 0.135 M NaOH that is required to neutralize 13.7 mL of 0.129 M HCl is 13.1 mL (option b).

Explanation:

The reaction between an acid and a base is called neutralization, forming a salt and water.

Salt is an ionic compound made up of an anion (positively charged ion) from the base and a cation (negatively charged ion) from the acid.

When an acid is neutralized, the amount of base added must equal the amount of acid initially present. This base quantity is said to be the equivalent quantity. In other words, at the equivalence point the stoichiometry of the reaction is exactly fulfilled (there are no limiting or excess reagents), therefore the numbers of moles of both will be in stoichiometric relationship. So:

V acid *M acid = V base *M base

where V represents the volume of solution and M the molar concentration of said solution.

In this case:

• V acid= 13.7 mL= 0.0137 L (being 1,000 mL= 1 L)
• M acid= 0.129 M
• V base= ?
• M base= 0.135 M

Replacing:

0.0137 L* 0.129 M= V base* 0.135 M

Solving:

$V base=\frac{0.0137 L*0.129 M}{0.135 M}$

V base=0.0131 L = 13.1 mL

The volume (mL) of 0.135 M NaOH that is required to neutralize 13.7 mL of 0.129 M HCl is 13.1 mL (option b).