Question

What volume of \(0.104 \mathrm{~N} \mathrm{H}_{2} \mathrm{SO}_{4}\) is required to neutralize \(15.2 \mathrm{~mL}\) of \(0.152 \mathrm{~N} \mathrm{NaOH}\) ? What volume of \(0.104 \mathrm{M} \mathrm{H}_{2} \mathrm{SO}_{4}\) is required to neutralize \(15.2 \mathrm{~mL}\) of \(0.152 \mathrm{M} \mathrm{NaOH} ?\) $$ \mathrm{H}_{2} \mathrm{SO}_{4}(a q)+2 \mathrm{NaOH}(a q) \rightarrow \mathrm{Na}_{2} \mathrm{SO}_{4}(a q)+2 \mathrm{H}_{2} \mathrm{O}(l) $$

Short answer

For both cases, approximately 11.0 mL of 0.104N (or 0.104M) H2SO4 is required to neutralize 15.2 mL of 0.152N (or 0.152M) NaOH.

Step-by-step solution

Case 1: Normality-based neutralization

For the case of normality-based neutralization, we'll utilize the relation: $$ N_\text{acid} × V_\text{acid} = N_\text{base} × V_\text{base} $$ Rearranging for the acid volume, we have: $$ V_\text{acid} = \frac{N_\text{base} × V_\text{base}}{N_\text{acid}} $$ Since one mole of H2SO4 reacts with two moles of NaOH to be neutralized, we need to correct acid normality. $$ N^\prime_\text{acid} = 2 × N_\text{acid} $$ Substituting the corrected N′acid into the equation and converting volumes into L: $$ V_\text{acid} = \frac{0.152\, \mathrm{N} \times 15.2\, \mathrm{mL} \times (1\,\mathrm{L} / 1000\,\mathrm{mL})}{2 \times 0.104\, \mathrm{N}} $$

Calculating Volume for Case 1

Now we can calculate the volume of the H2SO4 solution required to neutralize the given volume of NaOH solution for case 1: $$ V_\text{acid} = \frac{0.152 \times 15.2 \times (1/1000)}{2 \times 0.104} $$ $$ V_\text{acid} \approx 0.0110\,\mathrm{L} $$ Finally, converting the volume back to mL: $$ V_\text{acid} \approx 11.0\,\mathrm{mL} $$

Case 2: Molarity-based neutralization

For the case of molarity-based neutralization, the relation is similar, but we'll use molarities instead of normalities: $$ M_\text{acid} × V_\text{acid} = M_\text{base} × V_\text{base} $$ Rearranging for the acid volume, we have: $$ V_\text{acid} = \frac{M_\text{base} × V_\text{base}}{M_\text{acid}} $$ Since one mole of H2SO4 reacts with two moles of NaOH to be neutralized, we need to correct the acid molarity. $$ M^\prime_\text{acid} = 2 × M_\text{acid} $$ Substituting the corrected M′acid into the equation and converting volumes into L: $$ V_\text{acid} = \frac{0.152\, \mathrm{M} \times 15.2\, \mathrm{mL} \times (1\,\mathrm{L} / 1000\,\mathrm{mL})}{2 \times 0.104\, \mathrm{M}} $$

Calculating Volume for Case 2

Now we can calculate the volume of the H2SO4 solution required to neutralize the given volume of NaOH solution for case 2: $$ V_\text{acid} = \frac{0.152 \times 15.2 \times (1/1000)}{2 \times 0.104} $$ $$ V_\text{acid} \approx 0.0110\,\mathrm{L} $$ Finally, converting the volume back to mL: $$ V_\text{acid} = 11.0\,\mathrm{mL} $$ In this particular case, the volume of 0.104N (or 0.104M) H2SO4 required to neutralize 15.2 mL of 0.152N (or 0.152M) NaOH was found to be the same at around 11.0 mL.

Key concepts

Normality

Normality is a concept in chemistry that helps us understand the concentration of a solution in terms of equivalents of a solute per liter of solution. It is particularly useful in acid-base reactions, where it helps in calculating the amount needed to completely react with a given amount of another substance. Often abbreviated as "N," normality is a way to express the concentration that considers the reactivity of a substance.
In the case of acids and bases, normality takes into account how many hydrogen ions ( H^+ ) or hydroxide ions ( OH^− ) a substance can donate or accept in a reaction. This means that normality can be different from molarity, as it factors in the equivalents involved in the reaction. For example, sulfuric acid ( H_2SO_4 ), which can donate two hydrogen ions, will have a normality equal to twice its molarity in reactions where it fully dissociates.
When calculating neutralization reactions, such as the one involving H_2SO_4 and NaOH , the normality of H_2SO_4 is multiplied by two to reflect its ability to react with two equivalents of base. This correction is important to get accurate results for the volumes needed for neutralization.

Molarity

Molarity is one of the most common ways to express the concentration of a solution. It indicates how many moles of a solute are present per one liter of solution and is denoted by the symbol " M ." Understanding molarity is crucial for stoichiometry, as it helps in converting between volumes of reactants and products in a chemical reaction.
In simpler terms, molarity provides a direct relationship between the amount of a substance and the volume of liquid in which it is dissolved. For example, a 1 M (1 molar) solution of sulfuric acid ( H_2SO_4 ) is expected to have one mole of H_2SO_4 molecules in every liter of solution.
While molarity often suffices for many reactions, in acid-base reactions, such as the neutralization involving H_2SO_4 and NaOH , we must consider the reactive capacity of our acid. Since H_2SO_4 can release two hydrogen ions ( H^+ ), the reaction requires adjusting the molarity by multiplying it by two, similar to the adjustment in normality, to correctly reflect its full reactive capability in such reactions.

Acid-Base Reaction

An acid-base reaction is a fundamental concept in chemistry, describing the interaction between an acid and a base to form water and a salt. These reactions, also known as neutralization reactions, are key to many processes, both industrially and biologically.
In a typical acid-base reaction, the acid (H^+ donor) reacts with the base (OH^- donor) to produce water (H_2O), along with a compound called a salt. The general reaction can be summarized as:

• Acid + Base → Salt + Water

An example from the exercise is the reaction between sulfuric acid (H_2SO_4) and sodium hydroxide (NaOH), which produces sodium sulfate (Na_2SO_4) and water:
\[ H_2SO_4(aq) + 2NaOH(aq) → Na_2SO_4(aq) + 2H_2O(l) \] This reaction illustrates how the acid's hydrogen ions and the base's hydroxide ions combine to form water, highlighting the core concept of neutralization. The ratios of the reactants depend on their normality or molarity, ensuring that complete neutralization occurs.