Problem 79 - Chemistry An Atoms First Approach

Question

A student titrates an unknown amount of potassium hydrogen phthalate \(\left(\mathrm{KHC}_{8} \mathrm{H}_{4} \mathrm{O}_{4},\) often abbreviated KHP) with \right. \(20.46 \mathrm{mL}\) of a \(0.1000-M \mathrm{NaOH}\) solution. KHP (molar mass \(=\) \(204.22 \mathrm{g} / \mathrm{mol}\) ) has one acidic hydrogen. What mass of KHP was titrated (reacted completely) by the sodium hydroxide solution?

Step-by-Step Solution

Answer

The mass of KHP titrated (reacted completely) by the sodium hydroxide solution is approximately \(0.418 \mathrm{g}\).

Step 1: Calculate moles of NaOH used in the titration

First, we will calculate the moles of NaOH used in the titration using the given volume and molarity. moles of NaOH = volume of NaOH (in L) × molarity of NaOH moles of NaOH = \(20.46 \times 10^{-3} \mathrm{L} \times 0.1000 \frac{\mathrm{mol}}{\mathrm{L}}\)

Step 2: Determine moles of KHP that reacted

Since the reaction between KHP and NaOH is in a 1:1 ratio, the moles of KHP reacted will be equal to the moles of NaOH used in the titration. moles of KHP = moles of NaOH

Step 3: Calculate the mass of KHP titrated

Now, using the molar mass of KHP, we can determine the mass of KHP titrated. mass of KHP = moles of KHP × molar mass of KHP mass of KHP = moles of KHP × \(204.22 \frac{\mathrm{g}}{\mathrm{mol}}\)

Step 4: Calculate the final results

Plug the values from previous steps into the equations and solve for mass of KHP. moles of NaOH = \(20.46 \times 10^{-3} \mathrm{L} \times 0.1000 \frac{\mathrm{mol}}{\mathrm{L}} = 2.046 \times 10^{-3}\) mol moles of KHP = moles of NaOH = \(2.046 \times 10^{-3}\) mol mass of KHP = moles of KHP × \(204.22 \frac{\mathrm{g}}{\mathrm{mol}} = 2.046 \times 10^{-3} \mathrm{mol} \times 204.22 \frac{\mathrm{g}}{\mathrm{mol}} \approx 0.418 \mathrm{g}\) The mass of KHP titrated (reacted completely) by the sodium hydroxide solution is approximately 0.418 g.

Key Concepts

Molarity Calculation

Molarity is a key concept in chemistry that refers to the concentration of a solute in a solution. It is defined as the number of moles of solute dissolved in one liter of solution. To calculate molarity (M), you can use the formula:\[ \text{Molarity} (M) = \frac{\text{moles of solute}}{\text{liters of solution}} \]In the original exercise, the student uses a sodium hydroxide (NaOH) solution with a molarity of 0.1000 M.
This means that there are 0.1000 moles of NaOH in every liter of the solution. In order to find how many moles of NaOH were used, the student multiplies the volume of solution used (in liters) by the molarity:

• Volume of NaOH used: 20.46 mL, which is 20.46 × 10^-3 L
• Molarity of NaOH: 0.1000 M

Multiplying these values gives the number of moles of NaOH, allowing the student to proceed to stoichiometry calculations.

Stoichiometry

Stoichiometry involves calculating the relationships between reactants and products in a chemical reaction. It is a logical technique that uses the balanced chemical equation to find out the relative quantities of substances involved in the reaction.
In titrations, such as the one in the exercise, stoichiometry helps us understand the natural stoichiometric relationships between the reactants.In our example with KHP and NaOH, the balanced chemical equation shows a 1:1 molar ratio:\( \text{KHC}_8\text{H}_4\text{O}_4 (aq) + \text{NaOH} (aq) \rightarrow \text{KNaC}_8\text{H}_4\text{O}_4 (aq) + \text{H}_2\text{O} (l) \)This means that one mole of KHP reacts with one mole of NaOH. Therefore, the moles of KHP that have reacted will equal the moles of NaOH used. This can be shown from the earlier calculation of NaOH moles. Since we already know the moles of NaOH, these are directly equal to the moles of KHP, simplifying all subsequent mass calculations.

Chemical Reactions

Chemical reactions are processes where reactants transform into products. In the case of the titration between KHP and NaOH, the chemical reaction involves an acid-base neutralization.
When KHP (an acid) reacts with NaOH (a base), they combine to form water and a salt, as shown in the chemical equation. These types of reactions are fundamental in chemistry, because they balance according to the stoichiometry discussed earlier. The reaction of KHP and NaOH is a perfect example of a complete reaction.
By complete, we mean that all the reactant (KHP) is used up when it reacts with the NaOH provided. Given that the reaction follows a 1:1 stoichiometric ratio, all the sodium hydroxide is perfectly balanced by the KHP.
This thorough reaction allows us to determine exactly how much KHP was initially present, which the student does by examining the mass calculated from the moles of KHP, showing a full understanding and utility of chemical equations.