Question

Volume (in ml) of $0.7M$ $NaOH$ required for complete reaction with 350ml of $0.3M$ ${H_3}P{O_3}$ solution is:
A.$300ml$
B.$450ml$
C.$150ml$
D.$350ml$

Answer 1

By the concept of equivalent, ${n_1}{M_1}{V_1} = {n_2}{M_2}{V_2}$ .The n-factor of $NaOH = 1$ and the n-factor of ${H_3}P{O_3} = 2$ . Substitute all the values in this equation and calculate the solution.

Complete answer:
For solving this problem, we need to use the concept of equivalence.
The quantity of a substance which reacts with (or is equivalent to) an arbitrary quantity of another substance in a given chemical reaction is an equivalent (symbol: officially Equiv; unofficially but sometimes Eq). It is an obsolete unit of measurement that has been used in biological sciences and chemistry. Its relative weight is called the mass of an equivalent.
Or
The number of moles of an ion in a solution, multiplied by the value of that ion, is an equivalent.
In very simple words an equivalent gives us a relation between two substances that can react completely with each other.
Now, by the concept of equivalent
Equivalent $NaOH =$ Equivalent ${H_3}P{O_3}$
Since Equivalent $= NV$
Here, $N =$ Normality
$V =$ Volume
Again, since $N = nM$
Here, $n =$ n-factor
$M =$ Molarity
So, Equivalent $= nMV$
So, the initial equation becomes
${n_1}{M_1}{V_1} = {n_2}{M_2}{V_2}$
Here, ${n_1} =$ the n-factor of $NaOH = 1$
${M_1} =$ Molarity of $NaOH = 0.7M$
${V_1} =$ The volume of $NaOH$ that reacts with ${H_3}P{O_3}$
${n_2} =$ n-factor of ${H_3}P{O_3} = 2$
${M_2} =$ Molarity of ${H_3}P{O_3} = 0.3M$
${V_2} =$ Volume of ${H_3}P{O_3} = 350ml$
So, $1 \times 0.7 \times {V_1} = 0.3 \times 2 \times 350$
${V_1} = 300ml$
Hence, $300ml$ of $0.7M$ $NaOH$ is required for a complete reaction with 350ml of $0.3M$ ${H_3}P{O_3}$ solution.

Hence, option A is the correct choice.

Note:
Molarity - Molar concentration is a measure of the concentration of a chemical species, in particular of a solute in a solution. Normality – A way to measure solvent concentration is normality (N). It is similar to molarity, but in its expression of the solute quantity in a liter (L) of the solution, it uses the gram-equivalent weight of a liquid rather than the gram molecular weight expressed in molarity.