Question

Calculate the volume of \(1.50 \times 10^{-2} M \mathrm{NaOH}\) that must be added to \(500.0 \mathrm{~mL}\) of \(0.200 M \mathrm{HCl}\) to give a solution that has \(\mathrm{pH}=2.15\)

Short answer

To achieve a pH of 2.15 in the given solution, approximately 17.5 mL of \( 1.50 \times 10^{-2} M \) NaOH must be added to the 500.0 mL of 0.200 M HCl.

Step-by-step solution

Calculate moles of HCl

To calculate the initial moles of HCl in the solution, use the volume of the solution and the concentration of HCl: moles of HCl = volume × concentration moles of HCl = (500.0 mL) × (0.200 mol/L) Remember to convert mL to L: moles of HCl = (0.500 L) × (0.200 mol/L) = 0.100 mol #Step 2: Calculate the concentration of H_3O^+ ions from the desired pH#

Calculate [H_3O^+] from pH

pH is defined as the negative logarithm of the concentration of hydronium ions, H_3O^+. So, we can calculate the concentration of hydronium ions in the final solution using the given pH value: \[ [H_3O^+] = 10^{-pH} \] \[ [H_3O^+] = 10^{-2.15} \] \[ [H_3O^+] = 7.079 \times 10^{-3} \, M \] #Step 3: Calculate moles of H_3O^+ and HCl left in the solution#

Calculate moles of H_3O^+ and HCl

Since the target [H_3O^+] in the solution is given, we can find the moles of H_3O^+ in the solution: \[ \text{moles of H}_3\text{O}^{+} = (\text{total volume of solution} ) \times (\text{[H}_3\text{O}^{+}]) \] \[ \text{moles of H}_3\text{O}^{+} = (0.500 L + V_{\text{NaOH}}) \times (7.079 \times 10^{-3}) \] We also know that moles of HCl - moles of NaOH = moles of H_3O^+, Hence, moles of NaOH = moles of HCl - moles of H_3O^+. #Step 4: Calculate moles and volume of NaOH needed#

Calculate moles and volume of NaOH

Calculate the moles of NaOH required: moles of NaOH = 0.100 - (0.500 + V_{NaOH}) × (7.079 × 10^{-3} mol/L) Now, we can determine the volume of the NaOH solution: moles of NaOH = (volume of NaOH) × (concentration of NaOH) (0.100 - (0.500 + V_{NaOH}) × (7.079 × 10^{-3})) = V_{NaOH} × (1.50 × 10^{-2} mol/L) Solve for V_{NaOH} to obtain the volume of NaOH required: \[ V_{\text{NaOH}} = \frac{0.100 - (0.500 + V_{\text{NaOH}}) \times (7.079 \times 10^{-3})}{1.50 \times 10^{-2}} \] \[ V_{\text{NaOH}} \approx 0.0175 \, \text{L} \] So, the volume of \( 1.50 \times 10^{-2} M \) NaOH needed to achieve a pH of 2.15 is approximately 0.0175 L or 17.5 mL.

Key concepts

pH Calculation

Understanding how to calculate pH is key in acid-base titrations. pH measures the acidity or alkalinity of a solution. It's calculated using the formula:

This means you take the logarithm of the hydronium ion concentration and multiply by -1.
In this context, we calculated the pH to determine how much base is needed to reach a certain acidity (pH 2.15). Converting this pH to a hydronium ion concentration gives \([\text{H}_3\text{O}^+] = 7.079 \times 10^{-3} \text{ M}\). This value is crucial for adjusting the solution's acidity.

Molarity

Molarity is the concentration of a solution expressed as moles of solute per liter of solution.
It's calculated using the formula:

• Molarity (M) = \(\frac{\text{moles of solute}}{\text{volume of solution in liters}}\)

In the exercise, the molarity of HCl and NaOH were given as 0.200 M and 1.50 \(\times 10^{-2}\) M respectively.
Knowing these helps us calculate how much NaOH is needed to reach the desired pH. We used it to understand the balance between the amount of base added and the resulting concentration of HCl.

Stoichiometry

Stoichiometry involves using balanced chemical equations to calculate the quantities of reactants and products.
For acid-base reactions, like the one in this exercise, it helps determine how much of each substance reacts.

• Reaction: \(\text{HCl} + \text{NaOH} \rightarrow \text{NaCl} + \text{H}_2\text{O}\)

This reaction shows a 1:1 mole ratio between HCl and NaOH. We calculated moles left using the formula:

• Moles of NaOH = Moles of HCl - Moles of \(\text{H}_3\text{O}^+\)

This balance allowed us to find the exact amount of base to neutralize some excess acid.

Solution Volume

Solution volume is the total amount of space the liquids occupy in titrations. It's important because concentration calculations depend on it.
We often express volume in liters (L) or milliliters (mL). In this problem, we converted 500 mL of HCl to 0.500 L for our calculations.

• Total solution volume determines the dilution of ions in the solution.
• In the calculation of \([\text{H}_3\text{O}^+]\), the change in total volume includes the NaOH added.
• This impacts the final pH and concentration of the solution.

Understanding these changes in volume is crucial to ensure accurate chemical calculations and results.