Question

What volume of $$0.125M$$${H_2}S{O_4}$is required to neutralize$2.5g$of$Ca{(OH)_2}$?

Answer 1

Hint : In order to solve this, we need to write down it’s reaction first and then further we will use a formula to find out the volume required for the neutralization purpose. Neutralization is basically known as a chemical reaction in which acid and base reacts together to form salt, this is known as neutralization.

Complete Step By Step Answer:
Now, first we will write down a chemical reaction so that we can find out the no. of moles easily.
${H_2}S{O_{4(aqu)}} + Ca{(OH)_{2(aqu)}} \to CaS{O_{4(s)}} + 2{H_2}{O_{(l)}}$
Now, we will find out the molecular mass of $Ca{(OH)_2}$by adding up the mass of all the atoms present in the formula:
We know the mass of$Ca$, $O$, $H$are $14$, $16$, $1$.
Molecular mass $= 40 + (16 \times 2) + (1 \times 2) = 74gmo{l^{ - 1}}$
Since, we have the molecular mass, so we can find out the no. of moles present using this formula,
$n = \dfrac{m}{M}$, where n, m and M are number of moles, mass and molecular mass.
After putting the values into this formula we will find out the no. of moles of $Ca{(OH)_2}$
Therefore, $n = \dfrac{{2.5}}{{74}} = 0.0337mol$
Now, we have the number of moles of $Ca{(OH)_2}$present in the reaction and from the above reaction we also know that both the reactants are in $1:1$ratio. Therefore, we can say that we need one equivalent of ${H_2}S{O_4}$for the reaction to complete.
So, now we will divide the number of moles of $Ca{(OH)_2}$, which is, $0.0337mol$with the above-given molarity of ${H_2}S{O_4}$.
Therefore, volume of ${H_2}S{O_4}$required for neutralization is :
$= \dfrac{{0.0337mol}}{{0.125mol{L^{ - 1}}}} \times 1000mL \cdot {L^{ - 1}} = 270mL$
Hence, $270mL$ of $$0.125M$$${H_2}S{O_4}$is required to neutralize$2.5g$of$Ca{(OH)_2}$.

Note :
For solving such type of question we need to take care of the units as well. In this, we have to write down the reaction first because then only we will be able to find out the number of moles accurately and then only the further reaction will be done.