Question

# What mass of $\text {Mg (OH} {{\text {)}} _ {2}}$ is required to neutralize 25 ml of 0.136M hydrochloric acid solution? Mass of $\text {Mg (OH} {{\text {)}} _ {2}}$=58.33g/mol.(a) 0.248g(b) 0.496(c) 0.0992g(d) 1.98g

Hint: Using the formula of the molarity, the no of moles of the solute to the total volume of the solvent dissolved in 1l (1000ml) of the solution, first calculate the no of moles of HCl in 25 ml of 0.136M and then you can easily find the mass of $\text {Mg (OH} {{\text {)}} _ {2}}$ required. Solve it now.

$\text {Mg (OH} {{\text {)}} _ {2}}$ is basic in nature and HCl is acidic in nature. On reaction with each other, they undergo neutralization reaction i.e. that chemical reaction in which an acid and base reacts to form salt and water. The neutralization reaction of $\text {Mg (OH} {{\text {)}} _ {2}}$and HCl is as:
\begin{align} & \text{Mg(OH}{{\text{)}}_{2}}+2\text{HCl}\to \text{ Mg}{{\text{l}}_{2}}\text{+2}{{\text{H}}_{2}}\text{O} \\ & \text {base acid salt water} \\ \end{align} -----------(1)
From this it is clear that 2 moles of HCl require 58.33 g of $\text {Mg (OH} {{\text {)}} _ {2}}$. So, in order to find out the mass of $\text {Mg (OH} {{\text {)}} _ {2}}$ in 25 ml of 0.136M HCl, we should know the no of moles of HCl in it and that can be found by using the molarity equation as:

As we know that,
molarity =\dfrac {\text {no of moles of the solute}} {\begin{align} & \text {total volume of the solution} \\ & \text {dissolved in 1L (1000 ml) of the solution} \\ \end{align}}×
$\text{M}=\dfrac{\text{m}\times 1000} {\text{V}}$ ------------(2)
Where M is the molarity, m is the no of moles of solute i.e. HCl and V is the volume of the solution.
As M=0.136, V=25 (given)
Put all the values i.e. M and V in the above equation (2), we get:
$0.136=\dfrac{\text{m}\times 1000}{25}$
Now, from this we can easily find the value of no of moles of the HCl.
$\text{m}=\dfrac {0.136\times 25} {1000}$
m=0.0034
so, thus the 0.0034 are the moles of the HCl.
From the above equation (1), it is seen that the;
2 moles of HCl is neutralized by =58.33 g of $\text {Mg (OH} {{\text {)}} _ {2}}$
1mole of HCl is neutralized by = $\dfrac {58.33}{2}$
0.0034 moles of HCl is neutralized by=$\dfrac {\text {0} \text {.0034} \times 58.33}{2}$
=0.0992g of $\text {Mg (OH} {{\text {)}} _ {2}}$

∴0.0992g mass of $\text {Mg (OH} {{\text {)}} _ {2}}$is required to neutralize 25 ml of 0.136M hydrochloric acid solution.
So, the correct answer is “Option C”.

Note: Don’t confuse the words molarity and molality. Molarity depends on the volume of the solvent whereas the molality depends on the mass of the solvent. Molarity is as we know the no of moles of the solute to the total no of moles of the solvent dissolved in 1litre (1000ml) of the solution. On the other hand, the molality is the no of moles of the solute to the total mass of the solvent dissolved in 1kg (1000g) of the solution.