What volume of \(0.123 \mathrm{M} \mathrm{NaOH}(\mathrm{aq})\) must be added to \(125 \mathrm{~mL}\) of \(0.197 \mathrm{M} \mathrm{H}_{2} \mathrm{SO}_{3}(\mathrm{aq})\) to reach (a) the first stoichiometric point; (b) the second stoichiometric point?

Chemistry is like a dance of atoms and molecules. One of the fundamental steps in understanding chemical processes is learning how to write and balance chemical reaction equations. In an acid-base neutralization, such as the reaction between sodium hydroxide (NaOH) and sulfurous acid (H2SO3), it is essential to write the correct formulae for the reactants and products.

Let's take the exercise at hand. The first stoichiometric point occurs when one hydrogen ion from H2SO3 is neutralized. The balanced chemical equation for this step is:
\[ \text{H}_{2}\text{SO}_{3}(\text{aq}) + \text{NaOH}(\text{aq}) \rightarrow \text{NaHSO}_{3}(\text{aq}) + \text{H}_{2}\text{O}(\text{l}) \].
For the second stoichiometric point, the reaction involves the remaining hydrogen ion in NaHSO3. The balanced equation is:
\[ \text{NaHSO}_{3}(\text{aq}) + \text{NaOH}(\text{aq}) \rightarrow \text{Na}_{2}\text{SO}_{3}(\text{aq}) + \text{H}_{2}\text{O}(\text{l}) \].
Remember, writing balanced equations is crucial because it ensures that the law of conservation of mass is respected, indicating that the number of atoms of each element is the same on both sides of the equation.

Imagine if you could specify the strength of a potion. That's what chemists do with solutions using the concept of molarity! Molarity is just a way to talk about the concentration of a solution, indicating how many moles of a solute are in a liter of solution. It’s expressed in moles per liter (\( \text{mol/L} \) or \( \text{M} \)).

In this exercise, the strength or concentration of the sodium hydroxide (NaOH) solution is given as 0.123 M, which means there are 0.123 moles of NaOH in one liter of solution. Understanding this concept allows you to calculate the amount of reactant needed to achieve a full reaction - for example, to neutralize an acid in a titration.

Stoichiometry is the math behind chemistry. It tells us about the quantities of reactants and products involved in chemical reactions. Using the balanced equations and the molar concentrations, stoichiometric calculations allow us to predict how much of each substance will be needed or produced.

In our titration problem, we calculate the moles of H2SO3 present in 125 mL of 0.197 M solution. The calculation is simple: multiply the molarity of the H2SO3 by the volume it occupies (converted to liters, because molarity is per liter). You find that there are \(0.024625 \text{ mol} \) of H2SO3. Since the reaction ratios are 1:1 for both steps of the titration, you use the same number of moles of NaOH to reach the first and second stoichiometric points.

Acid-base titration is an analytical technique used to determine the concentration of an acid or a base by partially or fully neutralizing it with a base or an acid. Think of it as a game of ‘chemical balance’, where you add just enough titrant to react with the analyte until the equivalence point is reached.

Using the stoichiometry from the balanced equations, we can perform a titration calculation. For the first equivalence point, the reaction is a one-to-one ratio, meaning we use the same number of moles of NaOH calculated from H2SO3 (\(0.024625 \text{ mol} \)). Divide that by the molarity of the NaOH solution to find the volume required for neutralization. For the second equivalence point, we need twice as many moles of NaOH, since both hydrogen ions from H2SO3 need to be neutralized: \(0.024625 \text{ mol} \times 2 \). Thus, we find different volumes of NaOH needed for the first and second stoichiometric points, reflecting the stepwise neutralization in a diprotic acid titration.