91Ó°ÊÓ

A balanced chemical equation ensures the law of conservation of mass is followed, meaning that the number of atoms for each element is the same on both sides of the equation. For the titration between \(\text{H}_3\text{PO}_4\) and \(\text{NaOH}\), we need to recognize the elements and their respective counts. Phosphoric acid \(\text{H}_3\text{PO}_4\) reacts with sodium hydroxide \(\text{NaOH}\) to produce sodium phosphate \(\text{Na}_3\text{PO}_4\) and water \(\text{H}_2\text{O}\). By balancing the equation, we get: \[ \text{H}_3\text{PO}_4(aq) + 3\text{NaOH}(aq) \rightarrow \text{Na}_3\text{PO}_4(aq) + 3\text{H}_2\text{O}(l) \]In this equation, 1 mole of \(\text{H}_3\text{PO}_4\) reacts with 3 moles of \(\text{NaOH}\), forming 1 mole of \(\text{Na}_3\text{PO}_4\) and 3 moles of \(\text{H}_2\text{O}\). This balance is crucial for the molarity and mole ratio calculations.

Molarity (M) is a way to express the concentration of a solution and is defined as the number of moles of solute (the substance being dissolved) per liter of solution. The formula for molarity is written as: \[ M = \frac{\text{moles of solute}}{\text{liters of solution}} \]In our exercise, we are given the molarity of the \(\text{NaOH}\) solution as 0.204 M and the volume of \(\text{NaOH}\) solution used in titration, which is 16.4 mL(0.0164 L). Using the formula: \[ \text{Moles of NaOH} = \text{Molarity} \times \text{Volume} = 0.204 \times 0.0164 = 0.0033456 \]Next, we calculate the molarity of \(\text{H}_3\text{PO}_4\), for which we have the moles from the mole ratio step (0.0011152) and the total volume of \(\text{H}_3\text{PO}_4\) solution given (50.0 mL or 0.05 L). Therefore, we can find the molarity of \(\text{H}_3\text{PO}_4\): \[ M_{\text{H}_3\text{PO}_4} = \frac{0.0011152}{0.05} = 0.022304 \]

The mole ratio is derived from the coefficients of the reactants and products in a balanced chemical equation. It helps us convert between moles of different substances. From the balanced equation: \[ \text{H}_3\text{PO}_4(aq) + 3\text{NaOH}(aq) \rightarrow \text{Na}_3\text{PO}_4(aq) + 3\text{H}_2\text{O}(l) \]We see that 1 mole of \(\text{H}_3\text{PO}_4\) reacts with 3 moles of \(\text{NaOH}\). Therefore, to find the moles of \(\text{H}_3\text{PO}_4\) from the moles of \(\text{NaOH}\) calculated earlier (0.0033456 moles), we use the mole ratio: \[ \text{Moles of} \text{H}_3\text{PO}_4 = \frac{\text{Moles of NaOH}}{\text{Ratio factor}} = \frac{0.0033456}{3} = 0.0011152 \text{ moles} \]Understanding and using the mole ratio is crucial to relate quantities of reactants and products accurately in any chemical reaction.