What molarity of a 53.96 mL magnesium hydroxide solution is required to neutralize 93.24mL of a 7.306 M solution of hydrochloric acid solution, which creates as products, magnesium chloride and water?

Respuesta :

Answer

6.312 M

Explanation

Given:

Volume of Mg(OH)₂, Vb = 53.96 mL

Volume of HCl, Va = 93.24 mL

Molarity of HCl, Ma =7.306 M

What to find:

The molarity of Mg(OH)₂, Mb

Solution:

The first step is to write a balanced chemical equation for the reaction.

Mg(OH)₂ + 2HCl → MgCl₂ + 2H₂O

Mole ratio is 1:2; that is na = 2 and nb = 1

Hence, the molarity of Mg(OH)₂, Mb is calculated using the formula below:

[tex]\frac{M_aV_a}{n_a}=\frac{M_bV_b}{n_b}[/tex]

Plugging the values of the given parameters into the formula, we have:

[tex]\begin{gathered} \frac{7.306M\times93.24mL}{2}=\frac{M_b\times53.96mL}{1} \\ \\ Cross\text{ }multiply \\ \\ M_b\times53.96mL\times2=7.306M\times93.24mL\times1 \\ \\ Divide\text{ }both\text{ }sides\text{ }by\text{ }53.96mL\times2 \\ \\ \frac{M_b\times53.96mL\times2}{53.96mL\times2}=\frac{7.306M\times93.24mL\times1}{53.96mL\times2} \\ \\ M_b=6.312\text{ }M \end{gathered}[/tex]

Therefore, the molarity of a 53.96 mL magnesium hydroxide solution that is required to neutralize 93.24mL of a 7.306 M solution of the hydrochloric acid solution is 6.312 M