Question

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?

Answer 1

6.312 M

Step-by-step 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:

`(M_aV_a)/(n_a)=(M_bV_b)/(n_b)`

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

`\begin{gathered} (7.306M*93.24mL)/(2)=(M_b*53.96mL)/(1) \\ \\ Cross\text{ }multiply \\ \\ M_b*53.96mL*2=7.306M*93.24mL*1 \\ \\ Divide\text{ }both\text{ }sides\text{ }by\text{ }53.96mL*2 \\ \\ (M_b*53.96mL*2)/(53.96mL*2)=(7.306M*93.24mL*1)/(53.96mL*2) \\ \\ M_b=6.312\text{ }M \end{gathered}`

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