Question

Which of the following will have nearly equal \(\mathrm{H}^{-}\) concentration? (a) \(100 \mathrm{ml} 0.1 \mathrm{M}\) HCl mixed with \(50 \mathrm{ml}\) water (b) \(50 \mathrm{ml} 0.1 \mathrm{M} \mathrm{II}_{2} \mathrm{SO}_{4}\) mixed with \(50 \mathrm{ml}\) water (c) \(50 \mathrm{ml} 0.1 \mathrm{M} \mathrm{II}_{2} \mathrm{SO}_{4}\) mixed with \(100 \mathrm{ml}\) water (d) \(50 \mathrm{ml} 0.1 \mathrm{M}\) IICl mixcd with \(50 \mathrm{ml}\) watcr (1) a, b (2) \(\mathrm{b}, \mathrm{c}\) (3) \(\mathrm{a}, \mathrm{c}\) (4) \(\mathrm{c}, \mathrm{d}\)

Short answer

Options (a) and (c).

Step-by-step solution

Understanding the Problem

The problem requires identifying which options result in solutions with nearly equal \(\text{H}^{+}\) concentrations after mixing with water.

Calculate \( \text{H}^{+} \) Concentration for Option (a)

For option (a), 100 ml of 0.1 M HCl is mixed with 50 ml of water. The concentration of \( \text{H}^{+} \) ions can be calculated as follows: \[ \text{Concentration} = \frac{0.1 \text{ M} \times 100 \text{ ml}}{150 \text{ ml}} = \frac{10}{15} = 0.0667 \text{ M} \]

Calculate \( \text{H}^{+} \) Concentration for Option (b)

For option (b), 50 ml of 0.1 M H\textsubscript{2}SO\textsubscript{4} is mixed with 50 ml of water. The concentration of \( \text{H}^{+} \) ions is: \[ \text{Concentration} = \frac{0.2 \text{ M} \times 50 \text{ ml}}{100 \text{ ml}} = 0.1 \text{ M} \]

Calculate \( \text{H}^{+} \) Concentration for Option (c)

For option (c), 50 ml of 0.1 M H\textsubscript{2}SO\textsubscript{4} is mixed with 100 ml of water. The concentration of \( \text{H}^{+} \) ions is: \[ \text{Concentration} = \frac{0.2 \text{ M} \times 50 \text{ ml}}{150 \text{ ml}} = \frac{10}{150} = 0.0667 \text{ M} \]

Calculate \( \text{H}^{+} \) Concentration for Option (d)

For option (d), 50 ml of 0.1 M HCl is mixed with 50 ml of water. The concentration of \( \text{H}^{+} \) ions is: \[ \text{Concentration} = \frac{0.1 \text{ M} \times 50 \text{ ml}}{100 \text{ ml}} = 0.05 \text{ M} \]

Compare Concentrations

Compare the concentrations calculated in steps 2-5. The concentrations for options (a) and (c) are both 0.0667 M.

Key concepts

dilution formula

Understanding the dilution formula is key when working with solutions. The formula helps in calculating the resulting concentration of a solute after it has been diluted. The formula is: \[ C_1V_1 = C_2V_2 \] Where:

• C\textsubscript{1} is the initial concentration.
• V\textsubscript{1} is the initial volume.
• C\textsubscript{2} is the final concentration.
• V\textsubscript{2} is the final volume.

In the given exercise, various solutions of acids are being diluted by adding water. By carefully applying the dilution formula for each scenario, we can find the final concentrations of \(\text{H}^{+}\) ions.

molarity of acids

Molarity is a measure of the number of moles of solute per liter of solution. It is given by: \[ \text{Molarity} (M) = \frac{\text{moles of solute}}{\text{liters of solution}} \] For strong acids like HCl and H\textsubscript{2}SO\textsubscript{4}, this measure becomes crucial. In our exercise:

• Hydrochloric acid (HCl): It dissociates completely in water, so the molarity of HCl gives the exact concentration of \(\text{H}^{+}\) ions.
• Sulfuric acid (H\textsubscript{2}SO\textsubscript{4}): It is a diprotic acid, meaning it can donate two protons. Therefore, each mole of H\textsubscript{2}SO\textsubscript{4} dissociates to give two moles of \(\text{H}^{+}\) ions, doubling the effective concentration compared to its molarity.

By calculating the new molarities after dilution, we can compare them to see which solutions have nearly equal \(\text{H}^{+}\) concentrations.

acid-base chemistry

In acid-base chemistry, understanding the behavior and characteristics of acids and bases is crucial. Key points include:

• Strong acids - Such as HCl and H\textsubscript{2}SO\textsubscript{4}, which completely dissociate in water. Their \(\text{H}^{+}\) ion concentrations directly influence the solution's pH.
• Dilution impact - Adding water to an acidic solution decreases the concentration of \(\text{H}^{+}\) ions because the ions are spread out over a larger volume.
• pH calculation - The pH of a solution is calculated using \(-\text{log}[\text{H}^{+}]\). Solutions with similar \(\text{H}^{+}\) ion concentrations will have similar pH values, making them comparable in terms of acidity.

In our exercise, we see these principles in action. The mixed solutions demonstrate how different acids and their dilutions result in varying \(\text{H}^{+}\) concentrations, which are then compared to determine equivalence.