Problem 72 - Chemistry and Chemical Reactivity

Question

Potassium hydrogen phthalate, \(\mathrm{KHC}_{8} \mathrm{H}_{4} \mathrm{O}_{4},\) is used to standardize solutions of bases. The acidic anion reacts with strong bases according to the following net ionic equation: $$\mathrm{HC}_{8} \mathrm{H}_{4} \mathrm{O}_{4}^{-}(\mathrm{aq})+\mathrm{OH}^{-}(\mathrm{aq}) \rightarrow \mathrm{C}_{8} \mathrm{H}_{4} \mathrm{O}_{4}^{2-}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O}(\ell)$$ If a \(0.902-\mathrm{g}\) sample of potassium hydrogen phthalate is dissolved in water and titrated to the equivalence point with \(26.45 \mathrm{mL}\) of \(\mathrm{NaOH}(\mathrm{aq})\) what is the molar concentration of the NaOH?

Step-by-Step Solution

Answer

The molar concentration of NaOH is 0.167 M.

Step 1: Write the Balanced Reaction Equation

The provided net ionic equation is already balanced: \[\mathrm{HC}_{8} \mathrm{H}_{4} \mathrm{O}_{4}^{-} (\mathrm{aq}) + \mathrm{OH}^{-} (\mathrm{aq}) \rightarrow \mathrm{C}_{8} \mathrm{H}_{4} \mathrm{O}_{4}^{2-} (\mathrm{aq}) + \mathrm{H}_{2} \mathrm{O} (\ell)\]This indicates that one mole of hydroxide ion \((\mathrm{OH}^{-})\) reacts with one mole of potassium hydrogen phthalate ion \((\mathrm{HC}_{8} \mathrm{H}_{4} \mathrm{O}_{4}^{-})\).

Step 2: Calculate Moles of Potassium Hydrogen Phthalate

First, find the molar mass of potassium hydrogen phthalate, \(\mathrm{KHC}_{8}\mathrm{H}_{4}\mathrm{O}_{4}\):- \(\mathrm{K} = 39.10 \, \text{g/mol}\)- \(\mathrm{H} = 1.01 \, \text{g/mol} \times 5 = 5.05 \, \text{g/mol}\) (due to one \(\mathrm{H}^{+}\) in \(\mathrm{HKC}_{8}\mathrm{H}_{4}\mathrm{O}_{4}\))- \(\mathrm{C} = 12.01 \, \text{g/mol} \times 8 = 96.08 \, \text{g/mol}\)- \(\mathrm{O} = 16.00 \, \text{g/mol} \times 4 = 64.00 \, \text{g/mol}\)Adding these gives the total molar mass \(204.22 \, \text{g/mol}\).Calculate moles of potassium hydrogen phthalate:\[\text{Moles} = \frac{\text{Mass}}{\text{Molar Mass}} = \frac{0.902 \, \text{g}}{204.22 \, \text{g/mol}} = 0.00442 \, \text{mol}\]

Step 3: Use Stoichiometry to Find Moles of NaOH

From the balanced equation, we know 1 mole of \(\mathrm{KHC}_{8}\mathrm{H}_{4}\mathrm{O}_{4}^-\) reacts with 1 mole of \(\mathrm{OH}^-\). Therefore, the moles of \(\mathrm{NaOH}\) required are the same as moles of \(\mathrm{KHC}_{8}\mathrm{H}_{4}\mathrm{O}_{4}^-\) present, which is \(0.00442 \, \text{mol}\).

Step 4: Calculate Molarity of NaOH Solution

Molarity \((M)\) is given by the formula:\[M = \frac{\text{Moles of solute}}{\text{Volume of solution in liters}}\]Substitute the values:\(\text{Volume in liters} = \frac{26.45 \, \text{mL}}{1000} = 0.02645 \, \text{L}\)\(M = \frac{0.00442 \, \text{mol}}{0.02645 \, \text{L}} = 0.167 \, \text{M}\)

Key Concepts

Standardization of Solutions

Standardization is a crucial concept in chemistry, especially when it comes to solutions of acids and bases. It refers to the process of determining the exact concentration of a solution. This is typically done using a primary standard, which is a substance that is extremely pure, stable, non-hygroscopic, and has a high molecular weight. Potassium hydrogen phthalate (KHP) is often used as a primary standard for bases. This is because it is very pure and reacts predictably with strong bases like sodium hydroxide (NaOH).

In the standardization process, a known amount of KHP is dissolved in water and titrated with the base solution. The point at which the acid fully reacts with the base is known as the equivalence point. By knowing the amount of KHP that was used and the volume of the base added at equivalence, the concentration of the base can be calculated. This provides a precise molar concentration of the base solution, which is essential for accurate chemical reactions and analyses.

Molar Concentration Calculation

Molar concentration, or molarity, is a way of expressing the concentration of a solute in a solution. It is defined as the number of moles of a solute per liter of solvent. The formula to calculate molarity is given by:
\[ M = \frac{\text{Moles of solute}}{\text{Volume of solution in liters}} \]To determine the molar concentration of the NaOH solution in the given exercise, we first need to find the number of moles of the solute. In this case, the solute is NaOH. Next, we need the volume of the solution used during the titration. The volume must be converted from milliliters to liters for the calculation to be correct. Finally, by substituting the values into the molarity formula, we obtain the concentration. In this example, it was calculated to be 0.167 M, which means there are 0.167 moles of NaOH in one liter of the solution.

Net Ionic Equation

A net ionic equation provides a simplified way to look at chemical reactions, showing only the ions that directly participate in the reaction. In an acid-base titration, this approach helps to focus on the key substances that form new products.

The exercise involves a balanced net ionic equation:\[ \mathrm{HC}_{8} \mathrm{H}_{4} \mathrm{O}_{4}^{-} + \mathrm{OH}^{-} \rightarrow \mathrm{C}_{8} \mathrm{H}_{4} \mathrm{O}_{4}^{2-} + \mathrm{H}_{2} \mathrm{O} \]Here, the hydroxide ion (OH⁻) from the NaOH reacts with the acidic hydrogen phthalate ion (HC₈H₄O₄⁻) to produce phthalate ion (C₈H₄O₄²⁻) and water (H₂O). Understanding net ionic equations is fundamental as it highlights the essence of the chemical transformation rather than the auxiliary ions or molecules that do not change throughout the reaction.

Stoichiometry in Chemistry

Stoichiometry involves the calculation of reactants and products in chemical reactions. It is based on the balanced chemical equation, where coefficients indicate the number of moles of each substance involved. Stoichiometry allows chemists to predict the quantities of substances required for a reaction to proceed to completion.
In the provided exercise, stoichiometry is used to determine how many moles of NaOH are required to completely react with a given amount of potassium hydrogen phthalate. The balanced net ionic equation shows that one mole of HC₈H₄O₄⁻ reacts with one mole of OH⁻. Therefore, the moles of NaOH needed are equal to the moles of HC₈H₄O₄⁻ used, which is calculated from the mass and molar mass of the potassium hydrogen phthalate.

Using this relationship, we can accurately determine the molarity of the NaOH solution, which is essential for tasks such as solution preparation, reaction analyses, and quality control in chemical manufacturing.