Question

If $10{\text{ml}}$ of ${\text{1N HCl}}$ is mixed with $20{\text{mL}}$ of $1{\text{M }}{{\text{H}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}}$ and $30{\text{ml}}$of $1M{\text{ NaOH}}$. The resultant solution is:
A.20 meq of ${H^ + }$ ions
B.20 meq of $O{H^ - }$
C.0 meq of ${H^ + }$ or $O{H^ - }$
D.30 millimoles of ${H^ + }$

Answer 1

Hint: To solve this question, you must use the concept of equivalents. Equal numbers of equivalents of an acid neutralize equal numbers of equivalents of a base. Molarity of a solution when multiplied by its n factor gives the normality.
Formula used: m
${\text{meq}} = {{M \times V \times }}x$
Where ${\text{meq}}$ is the number of milliequivalents
M is the molarity of the used acid / base solution
V is the volume of the solution
x is the n-factor of the acid/ base used

Complete step by step answer:
The n factor of a compound is also known as the valence factor and its value depends upon the compound that is being considered. For instance, for an acid, the basicity of the acid is its n factor. Similarly, for a base, its basicity is taken as the n factor.
For $NaOH$, n- factor$ = 1$, for $HCl$ and ${H_2}S{O_4}$, the n factors are 1 and 2 respectively.
To find the concentration of ${H^ + }$ ions in the solution, we use the formula,
${\text{meq of acid }} - {\text{meq of base}} = {\text{meq of }}{{\text{H}}^ + }$
${N_{HCl}} \times {V_{HCl}} + {N_{{H_2}S{O_4}}} \times {V_{{H_2}S{O_4}}} - {N_{NaOH}} \times {V_{NaOH}} = {\text{meq of }}{{\text{H}}^ + }$
Substituting the values:
$ \Rightarrow (1 \times 10) + \left( {1 \times 2 \times 20} \right) - \left( {30 \times 1} \right) = 10 + 40 - 30$
$\therefore {\text{meq of }}{{\text{H}}^ + } = 20$

Thus, the correct answer is A.

Note:
Stoichiometry is based upon the very basic laws of chemistry that help to understand it better, namely, the law of conservation of mass, the law of definite proportions (the law of constant composition), the law of reciprocal proportions and the law of multiple proportions .In general, different chemicals combine in definite ratios in chemical reactions. Since matter can neither be created nor destroyed, thus the amount of each element must be the same throughout the entire reaction.