Question

Assuming complete dissociation of ${H_2}S{O_4}$ as $({H_2}S{O_4} + 2{H_2}O \to 2{H_2}{O^ + } + SO_4^{2 - })$. The number of ${H_3}{O^ + }$ ions in $10mL$ of $0.5N\;{H_2}S{O_4}$ is
(A) $3.01 \times {10^{21}}$
(B) $1.5 \times {10^{21}}$
(C) $6.02 \times {10^{21}}$
(D) $3.01 \times {10^{22}}$

Answer 1

We know that normality of a solution is defined as the moles of solute multiplied by n factor and divided by volume of solute, using this condition we can first identify the moles of sulphuric acid and with those moles we can easily calculate the number of hydronium ions.

Complete Step by step answer: As we know that normality is defined as the number of gram equivalents of solute present in per litre of solution and when weight of solute is not given then we can simply replace the quantity with moles so normality can also be defined as the moles of solute multiplied by n-factor and dividing them by the given volume. We can give this formula as:
$N = \dfrac{{moles\;of\;solute \times n \times 1000}}{{volume(ml)}}$

Using this formula we can calculate the number of moles of sulphuric acid (${H_2}S{O_4}$):
$moles = \dfrac{{N \times vol.}}{{n \times 1000}}$
$\Rightarrow moles = \dfrac{{0.5 \times 10}}{{2 \times 1000}}$
$\Rightarrow moles = 2.5 \times {10^{ - 3}}mol$
Now from the given equation of complete dissociation of sulphuric acid we can see that $1$ mole of ${H_2}S{O_4}$ reacts with $2$ mole of water, so $2.5 \times {10^{ - 3}}$ moles of ${H_2}S{O_4}$ will react with $2 \times 2.5 \times {10^{ - 3}}$ moles of hydronium ions. So moles of ${H_3}{O^ + } = 5 \times {10^{ - 3}}mol$.
Now, using these moles of hydronium ion we can calculate the number of hydronium ions as:
$moles = \dfrac{{no.\;of\;molecules}}{{{N_A}}}$
$\Rightarrow no.\;of\,molecules = 5 \times {10^{ - 3}} \times 6.022 \times {10^{23}}$
$\Rightarrow no.\;of\,molecules = 3.01 \times {10^{21}}$

Therefore we can say that the correct answer is (A).

Additional information: if we are given with normality in the question and asked to solve for molarity we can calculate it directly by dividing the normality of the given solution with the n-factor of the solute.

Note: Remember that complete dissociation means the complete separation of molecules into its constituent ions or atoms or radicals. Also the n-factor of sulphuric acid is given as the number of hydrogen ions replaced by $1$ mole of acid in the reaction. As sulphuric acid possesses $2$ replaceable hydrogen atoms in this case so n-factor is equal to $2$.