Question

Which of the following solutions will have \(\mathrm{pH}\) close to \(1.0 ?\) (a) \(100 \mathrm{ml}\) of \((\mathrm{M} / 10) \mathrm{HCl}+100 \mathrm{ml}\) of \((\mathrm{M} / 10) \mathrm{NaOH}\) (b) \(55 \mathrm{ml}\) of \((\mathrm{M} / 10) \mathrm{HCl}+45 \mathrm{ml}\) of \((\mathrm{M} / 10) \mathrm{NaOH}\) (c) \(10 \mathrm{ml}\) of \((\mathrm{M} / 10) \mathrm{HCl}+90 \mathrm{ml}\) of \((\mathrm{M} / 10) \mathrm{NaOH}\) (d) \(75 \mathrm{ml}\) of \((\mathrm{M} / 10) \mathrm{HCl}+25 \mathrm{ml}\) of \((\mathrm{M} / 5) \mathrm{NaOH}\)

Short answer

Option (d) will have a pH closest to 1.0, at approximately 1.6.

Step-by-step solution

Calculate moles of HCl and NaOH for each option

For each option, we need to calculate the moles of HCl and NaOH present by using the formula for moles: \( ext{moles} = ext{concentration (M)} \times ext{volume (L)} \).(a) - HCl: \( 100 \text{ mL} = 0.1 \text{ L} \) and concentration is \( \frac{M}{10} = 0.1 \text{ M} \) \[ ext{Moles of HCl} = 0.1 \times 0.1 = 0.01 \text{ moles} \] - NaOH: \( 100 \text{ mL} = 0.1 \text{ L} \) and concentration is \( \frac{M}{10} = 0.1 \text{ M} \) \[ ext{Moles of NaOH} = 0.1 \times 0.1 = 0.01 \text{ moles} \](b) - HCl: \( 55 \text{ mL} = 0.055 \text{ L} \) and concentration is \( \frac{M}{10} = 0.1 \text{ M} \) \[ ext{Moles of HCl} = 0.1 \times 0.055 = 0.0055 \text{ moles} \] - NaOH: \( 45 \text{ mL} = 0.045 \text{ L} \) and concentration is \( \frac{M}{10} = 0.1 \text{ M} \) \[ ext{Moles of NaOH} = 0.1 \times 0.045 = 0.0045 \text{ moles} \] (c) - HCl: \( 10 \text{ mL} = 0.01 \text{ L} \) and concentration is \( \frac{M}{10} = 0.1 \text{ M} \) \[ ext{Moles of HCl} = 0.1 \times 0.01 = 0.001 \text{ moles} \] - NaOH: \( 90 \text{ mL} = 0.09 \text{ L} \) and concentration is \( \frac{M}{10} = 0.1 \text{ M} \) \[ ext{Moles of NaOH} = 0.1 \times 0.09 = 0.009 \text{ moles} \](d) - HCl: \( 75 \text{ mL} = 0.075 \text{ L} \) and concentration is \( \frac{M}{10} = 0.1 \text{ M} \) \[ ext{Moles of HCl} = 0.1 \times 0.075 = 0.0075 \text{ moles} \] - NaOH: \( 25 \text{ mL} = 0.025 \text{ L} \) and concentration is \( \frac{M}{5} = 0.2 \text{ M} \) \[ ext{Moles of NaOH} = 0.2 \times 0.025 = 0.005 \text{ moles} \]

Determine excess moles of HCl

We need to find which option has un-reacted HCl left over, since that will contribute to the pH.(a) The moles of HCl and NaOH are equal (0.01 moles each), so they will neutralize each other. (b) Moles of HCl = 0.0055 and moles of NaOH = 0.0045, giving an excess of 0.001 moles of HCl because \[ 0.0055 - 0.0045 = 0.001 \text{ moles of HCl} \] (c) Moles of HCl = 0.001 and moles of NaOH = 0.009, NaOH is in excess so no HCl remains.(d) Moles of HCl = 0.0075 and moles of NaOH = 0.005, giving an excess of \[ 0.0075 - 0.005 = 0.0025 \text{ moles of HCl} \]

Calculate resulting pH

The pH is calculated using the concentration of H+ ions from the excess HCl. The formula for pH is: \[ \text{pH} = -\log_{10} [\text{H}^+] \]. Calculate [H+] by dividing the excess moles of HCl by the total volume in liters.(b) The total volume is \( 100\text{ mL} = 0.1\text{ L} \). \[ [\text{H}^+] = \frac{0.001}{0.1} = 0.01 \text{ M} \]Thus, \[ \text{pH} = -\log_{10}(0.01) = 2 \](d) The total volume is \( 100\text{ mL} = 0.1\text{ L} \). \[ [\text{H}^+] = \frac{0.0025}{0.1} = 0.025 \text{ M} \]Thus,\[ \text{pH} = -\log_{10}(0.025) \approx 1.6 \]

Determine the solution with pH closest to 1.0

Option (d) has a pH of approximately 1.6 which is the closest to a pH of 1.0 compared to the others.

Key concepts

Acid-Base Reactions

Acid-base reactions are an exciting area of chemistry where acids and bases interact to neutralize each other. Here's how it works: when an acid, like Hydrochloric acid (HCl), meets a base, such as Sodium hydroxide (NaOH), they react to form water and a salt. This process releases heat. In our exercise, each solution combines varying amounts of HCl and NaOH, which helps in determining which solutions are fully neutralized or have excess acid left. Unlike other types of reactions, acid-base reactions involve the transfer of protons (H⁺ ions). The acid donates protons, while the base accepts them. This proton transfer is what leads to neutralization and is central to calculating the resulting pH after the reaction. While calculating pH, remember that if HCl (acid) is in excess after the reaction, the solution will be more acidic, thus having a lower pH. Conversely, if there's more NaOH left, the solution becomes more basic, yielding a higher pH.

Mole Calculations

Moles are like the bridge between the world of atoms and the substances we directly interact with. In chemistry, the concept of "moles" helps quantify substances precisely. The key to solving our exercise lies within calculating the number of moles for both HCl and NaOH.To find the number of moles, use the formula: - \[ \text{Moles} = \text{Concentration (M)} \times \text{Volume (L)} \] - This formula ensures accuracy when dealing with varying solution strengths and volumes, allowing us to compare the substances directly.For example, to determine the moles of HCl in the first scenario, convert the milliliters to liters (100 mL = 0.1 L), multiply by the given molarity (0.1 M), and voila! You have your moles of HCl. Repeating the same for NaOH or any other solutions allows for perfect comparisons, crucial in deciding which component is in excess after a reaction. This step-by-step approach clarifies which solution will have an excess of H+ ions, thus affecting the pH.

Acid Concentration

Acid concentration plays a vital role in determining the solution's pH. The concentration tells us how many H⁺ ions are in a solution. The higher the concentration of these ions, the more acidic the solution becomes.In our problem, each scenario has a different mixture of acid (HCl) and base (NaOH). This leads to differences in the concentration of H+ ions left in the solution after reacting. When you know the excess moles of HCl, divide them by the total solution volume to get the concentration of hydrogen ions - \( [\text{H}^+] = \frac{\text{Excess moles of HCl}}{\text{Total volume (L)}} \).This concentration directly influences the pH. Given the logarithmic nature of the pH scale, even small changes in concentration can have significant effects on pH values. Understanding acid concentration is crucial to mastering acid-base chemistry and confidently calculating pH.

Neutralization

Neutralization is a fascinating process where an acid and base react to form water and a salt. It's a complete balancing act, where both acid and base cancel each other out. In essence, the number of H⁺ ions from the acid matches the OH⁻ ions from the base. In our exercise, various solutions are tested to see if they achieve neutralization. If the moles of acid equal the moles of base, the solution will be neutral. However, if one is more than the other, a remnant of that compound is left unreacted. Measuring neutralization helps predict the solution's pH. Fully neutralized solutions will have a pH near neutral, around 7. However, if there's remaining acid, the solution will show a pH lower than 7, indicating acidity. Keeping these principles in mind allows anyone to handle different mixtures accurately and understand the remaining acidity or basicity in a solution.