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 \(\mathrm{M} \mathrm{HCl}\) to give a solution that has \(\mathrm{pH}=2.15 .\)

Short answer

To obtain a pH of 2.15 in a 500.0 mL 0.200 M HCl solution, you need to add approximately 17.68 mL of \(1.50 \times 10^{-2}\) M NaOH solution.

Step-by-step solution

Calculate moles of HCl in the initial solution

To find the moles of HCl in the initial solution, we use the formula: moles = Molarity x Volume where Molarity = 0.200 M and Volume = 500.0 mL or 0.500 L (since 1 L = 1000 mL). moles = 0.200 M * 0.500 L = 0.100 moles of HCl Step 2: Calculate the moles of H+ ions after adding NaOH

Determine moles of H+ ions required for the desired pH

We use the pH formula to find the moles of H+ ions at a pH of 2.15: pH = -log[H+] 2.15 = -log[H+] By rewriting the equation, we find the concentration of H+ ions: [H+] = 10^{-2.15} = 7.079 x 10^{-3} M Now we calculate the moles of H+ ions in the solution after the neutralization reaction: moles of H+ ions after adding NaOH = (final concentration of H+ ions) x (final volume of solution) The final volume of the solution will be the initial volume of HCl solution (0.500 L) plus the volume of NaOH solution added. Step 3: Calculate the moles of NaOH required

Determine moles of NaOH needed to neutralize excess H+ ions

Since NaOH and HCl react in a 1:1 ratio, the moles of NaOH required will be equal to the moles of H+ ions neutralized: moles of H+ ions neutralized = initial moles of HCl - final moles of H+ ions Let x be the volume of NaOH solution to be added: x L * (1.50 x 10^{-2} M) = 0.100 moles HCl - [(7.079 x 10^{-3} M) * (0.500 L + x)] Step 4: Solve for the volume of NaOH solution

Calculate the volume of NaOH solution

Solve the equation above for the volume (x) of NaOH solution required: x L * (1.50 x 10^{-2} M) = 0.100 - [(7.079 x 10^{-3} M) * (0.500 + x)] Now, let's isolate x on one side of the equation: x (1.50 x 10^{-2} M + 7.079 x 10^{-3} M) = 0.100 - 7.079 x 10^{-3} M * 0.500 Divide both sides by the coefficients of x: x = (0.100 - [7.079 x 10^{-3} * 0.500])/(1.50 × 10^{-2} M + 7.079 x 10^{-3} M) Now we can calculate the value of x: x ≈ 0.01768 L To obtain the volume in milliliters, multiply this value by 1000: 0.01768 L * 1000 = 17.68 mL So, we need to add approximately 17.68 mL of 1.50 x 10^{-2} M NaOH to 500.0 mL of 0.200 M HCl solution to obtain a pH of 2.15.

Key concepts

Molarity Calculations

Molarity is a fundamental concept in chemistry that helps us quantify the concentration of a solution. It's defined as the number of moles of solute per liter of solution. To calculate molarity, use the formula: - Molarity (M) = moles of solute / liters of solution
In the context of acid-base titrations, it's crucial for determining how much of one solution is needed to react with another. For example, in the exercise, we have a solution of hydrochloric acid (HCl) with a molarity of 0.200 M and a volume of 500.0 mL. To calculate the moles of HCl present, you would convert the volume to liters and multiply by the molarity: - 500.0 mL = 0.500 L - Moles of HCl = 0.200 M * 0.500 L = 0.100 moles
Understanding molarity allows us to accurately gauge how much of a chemical is present in a solution, which is essential for further calculations in chemical reactions.

Neutralization Reaction

Neutralization is a chemical reaction between an acid and a base that results in the formation of water and a salt. In this exercise, sodium hydroxide (NaOH), a base, reacts with hydrochloric acid (HCl), an acid. The reaction is: - \[ ext{NaOH} + ext{HCl} ightarrow ext{NaCl} + ext{H}_2 ext{O} \]
This type of reaction typically occurs in a 1:1 molar ratio, meaning one mole of NaOH neutralizes one mole of HCl. Here, the goal is to use NaOH to reduce the concentration of H+ ions in the HCl solution, thus affecting the solution's pH.
The initial moles of H+ ions from HCl in this scenario were calculated as 0.100 moles. After deciding on the desired pH, we can calculate how much NaOH is needed to neutralize enough of these ions. Once the calculations are done, we can determine the volume of NaOH solution necessary to achieve the target pH by adding an exact amount of NaOH corresponding to the H+ ions required.

pH Calculation

pH is a measure of how acidic or basic a solution is. It is calculated using the concentration of hydrogen ions in a solution: - \[ ext{pH} = - ext{log}[ ext{H}^+] \]
In our problem, we're aiming for a final pH of 2.15. To find the concentration of hydrogen ions that corresponds to this pH, rearrange the equation: - \[ [ ext{H}^+] = 10^{- ext{pH}} \]
By substituting the desired pH into the formula, we get: - \[ [ ext{H}^+] = 10^{-2.15} \] - \[ [ ext{H}^+] ext{ or final concentration of H+} = 7.079 imes 10^{-3} ext{ M} \]
To conclude how many moles of H+ remain after neutralization, use this [H+] to adjust the volume of the acid-base mixture, ensuring the calculation fits the system's constraints.

Volume Conversion

Volume conversion is critical when dealing with solutions in chemistry, especially when the original measurements are in milliliters but calculations require liters. This conversion is essential for accurate molarity calculations and subsequent chemical reactions.
In the exercise, the initial volume of HCl is given as 500.0 mL. To convert this to liters, use the conversion factor: - 1 L = 1000 mL - Thus, \[ 500.0 ext{ mL} = 0.500 ext{ L} \]
Similarly, once the volume of NaOH required is calculated in liters (about 0.01768 L in this case), it’s often useful to convert it back to milliliters for practical lab measurements: - \[ 0.01768 ext{ L} imes 1000 = 17.68 ext{ mL} \]
Volume conversion ensures precision when measuring and mixing solutions during chemical reactions, facilitating accurate experimental outcomes.