Question

You have separate solutions of HCl and H2SO4 with the same concentrations in terms of molarity. You wish to neutralize a solution of NaOH. Which acid solution would require more volume (in mL) to neutralize the base? A. The HCl solution. B. The H2SO4 solution. C. You need to know the acid concentrations to answer this question. D. You need to know the volume and concentration of the NaOH solution to answer this question. E. C and D

Answer 1

Answer: Option (A) is the correct answer.

Explanation:

A neutralization reaction is defined as the reaction in which an acid reacts with a base to yield salt and water.

The given reaction will be as follows.

         $HCl + NaOH \rightarrow NaCl + H_{2}O$

       $H_{2}SO_{4} + 2NaOH \rightarrow Na_{2}SO_{4} + 2H_{2}O$

Now, if molarity of NaOH = $M_{1}$ and volume of NaOH = $V_{1}$

Therefore, Molarity of HCl = molarity of $H_{2}SO_{4}$ = $M_{2}$

     Volume of HCl required = $\text{molarity of NaOH} \times \frac{\text{volume of NaOH}}{\text{molarity of HCl}}$

                 = $\frac{M_{1}V_{1}}{M_{2}}$

So, vVolume of $H_{2}SO_{4}$ required = 0.5 x molarity of NaOH x \frac{\text{volume of NaOH}}{\text{molarity of H_{2}SO_{4}}}[/tex]

                       = $0.5 \times \frac{M_{1}V_{1}}{M_{2}}$

Hence, volume of HCl = 2 x volume of $H_{2}SO_{4}$.

Thus, we can conclude that HCl solution would be required more volume (in mL) to neutralize the base.