Question

In a titration of \(\mathrm{HNO}_{3}\), you add a few drops of phenolphthalein indicator to \(50.00 \mathrm{~mL}\) of acid in a flask. You quickly add \(20.00 \mathrm{~mL}\) of \(0.0502 \mathrm{M} \mathrm{NaOH}\) but overshoot the end point, and the solution turns deep pink. Instead of starting over, you add \(30.00 \mathrm{~mL}\) of the acid, and the solution turns colorless. Then, it takes \(3.22 \mathrm{~mL}\) of the \(\mathrm{NaOH}\) to reach the end point. (a) What is the concentration of the \(\mathrm{HNO}_{3}\) solution? (b) How many moles of \(\mathrm{NaOH}\) were in excess after the first addition?

Short answer

(a) The concentration of the \mathrm{HNO\_{3}} solution is 0.0468 M. (b) 0.000842 moles of NaOH were in excess after the first addition.

Step-by-step solution

Calculate moles of NaOH added

Use the formula for moles, which is \text{Molarity} \times \text{Volume}. Therefore, for the first addition: \[ n(\text{NaOH}) = 0.0502 \text{ M} \times 20.00 \text{ mL} \times \frac{1 \text{ L}}{1000 \text{ mL}} = 0.001004 \text{ moles} \]

Calculate moles of excess NaOH

Determine the moles of NaOH required to neutralize the added HNO3. Since it took 3.22 mL of NaOH to reach the endpoint after adding 30.00 mL of acid, the moles of NaOH required are: \[ n(\text{NaOH}_{\text{used}}) = 0.0502 \text{ M} \times 3.22 \text{ mL} \times \frac{1 \text{ L}}{1000 \text{ mL}} = 0.000161644 \text{ moles} \] Thus, the moles in excess after the first NaOH addition is \[ 0.001004 - 0.000161644 = 0.000842356 \text{ moles} \]

Calculate moles of HNO3 neutralized by excess NaOH

The 30.00 mL of HNO3 neutralizes the excess NaOH completely. So, the moles of HNO3 neutralized are equal to the moles of excess NaOH: \[ n(\text{HNO3}) = 0.000842356 \text{ moles} \]

Determine the initial concentration of HNO3

The total moles of HNO3 initially present in the 50.00 mL of solution must account for the 0.000842356 moles additionally required plus the moles of NaOH that react later. First, find moles neutralized by the second addition of NaOH: \[ 30.00 \text{ mL} \times C(\text{HNO3}) = 0.000842356 \text{ moles} \] Solve for C(HNO3): \[ C(\text{HNO3}) = \frac{0.000842356 \text{ moles}}{30.00 \text{ mL}} \times \frac{1000 \text{ mL}}{1 \text{ L}} = 0.02808 \text{ M} \]

Correct for initial solution volume

Multiply the calculated concentration by the ratio of the initial volume over the volume neutralized to get the correct concentration: \[ C(\text{HNO3})_{\text{initial}} = 0.02808 \text{ M} \times \frac{50.00 \text{ mL}}{30.00 \text{ mL}} = 0.0468 \text{ M} \]

Key concepts

Molarity Calculations

Molarity tells us the concentration of a solution. It is defined as the number of moles of solute dissolved in one liter of solution. The formula to calculate molarity is: Molarity (M) = \( \frac{\text{moles of solute}}{\text{volume of solution in liters}} \) Always remember to convert the volume into liters before using the formula. For instance, if you have added 20.00 mL of a solution with a molarity of 0.0502 M, you first convert 20.00 mL to liters (which is 0.020 L) and then multiply it by the molarity to find the number of moles: n(NaOH) = 0.0502 M \( \times \) 0.020 L = 0.001004 moles This way, you can find the number of moles present in any given volume of solution.

Neutralization Reaction

A neutralization reaction occurs when an acid reacts with a base to produce salt and water. The general formula for this reaction is: \( \text{Acid} + \text{Base} \rightarrow \text{Salt} + \text{Water} \) In the given exercise, NaOH and HNO3 neutralize each other. When you add sodium hydroxide (NaOH, a base) to nitric acid (HNO3, an acid), they react to form sodium nitrate (NaNO3) and water (H2O). When the phenolphthalein indicator changes color, it signals the endpoint of the titration, meaning the acid has been neutralized by the added base. Sometimes, overshooting the endpoint might happen, leading to an excess amount of base in the solution.

Acid-Base Titration

Titration is a technique used to determine the concentration of a solute in a solution. In an acid-base titration, a solution of known concentration (titrant) is added to a solution of unknown concentration (analyte) until the reaction reaches the endpoint. Some key points to remember:

• The equivalence point is when the number of moles of acid equals the number of moles of base.
• Indicators, like phenolphthalein, are used to show the endpoint of the titration visibly.
• In this exercise, NaOH is the titrant and HNO3 is the analyte.

The titration helps you find out the concentration of the HNO3 by calculating the amount of NaOH required to reach the endpoint.

Phenolphthalein Indicator

Phenolphthalein is a chemical compound used as an indicator in titrations. During a titration, it helps signal the endpoint of the reaction. Key characteristics of phenolphthalein:

• It is colorless in acidic solutions.
• It turns pink in basic solutions.
• The color change occurs around a pH of 8.2 to 10.

In this exercise, phenolphthalein changes color to deep pink when an excess of NaOH is added, indicating the solution is now basic. Adding more nitric acid (HNO3) then returns the solution to colorless, showing that the NaOH has been neutralized.

Stoichiometry

Stoichiometry involves using balanced chemical equations to calculate the quantities of reactants and products involved in a chemical reaction. It is essential for solving titration problems. In the titration exercise:

• The balanced equation for the reaction between NaOH and HNO3 is: \( \text{NaOH} + \text{HNO3} \rightarrow \text{NaNO3} + \text{H2O} \)
• Using the stoichiometric relationship, you can relate the moles of NaOH to the moles of HNO3.

From the titration data, you calculate the number of moles of NaOH used, then use stoichiometry to determine the number of moles and ultimately the concentration of HNO3 based on the volumes and molarities involved. Always ensure your chemical equations are balanced to apply stoichiometry correctly.