Problem 131 - A Complete Resource Book in Chemistry for JEE Main

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 \((\mathrm{M} / 5) \mathrm{NaOH}\)

Step-by-Step Solution

Answer

Option (d) has a pH closest to 1.

Step 1: Calculate Moles of HCl in Option (a)

In option (a), we have 100 mL of \((M/10)\) HCl. The concentration is \(0.1\ M\), so the moles of HCl are: \[ \text{Moles of HCl} = 100 \ ext{mL} \times \frac{1 \ ext{L}}{1000 \ ext{mL}} \times 0.1 \ M = 0.01 \ ext{mol} \]

Step 2: Calculate Moles of NaOH in Option (a)

In option (a), we also have 100 mL of \((M/10)\) NaOH with the same concentration of \(0.1\ M\). The moles of NaOH are: \[ \text{Moles of NaOH} = 100 \ ext{mL} \times \frac{1 \ ext{L}}{1000 \ ext{mL}} \times 0.1 \ M = 0.01 \ ext{mol} \]

Step 3: Determine Final Solution for Option (a)

In option (a), both the HCl and NaOH have equal moles, thus they will neutralize each other completely. The final solution will be neutral with a pH of 7.

Step 4: Calculate Moles of HCl in Option (b)

In option (b), we have 55 mL of \((M/10)\) HCl. The concentration is \(0.1\ M\), so the moles of HCl are: \[ \text{Moles of HCl} = 55 \ ext{mL} \times \frac{1 \ ext{L}}{1000 \ ext{mL}} \times 0.1 \ M = 0.0055 \ ext{mol} \]

Step 5: Calculate Moles of NaOH in Option (b)

In option (b), we have 45 mL of \((M/10)\) NaOH. The concentration is \(0.1\ M\), so the moles of NaOH are: \[ \text{Moles of NaOH} = 45 \ ext{mL} \times \frac{1 \ ext{L}}{1000 \ ext{mL}} \times 0.1 \ M = 0.0045 \ ext{mol} \]

Step 6: Determine Excess HCl in Option (b)

In option (b), the moles of HCl exceed those of NaOH, hence there is an excess of \(0.0055 - 0.0045 = 0.001\ \text{mol}\) of HCl left.

Step 7: Calculate pH for Option (b)

The pH of the resulting solution in option (b) is determined by the concentration of excess HCl. The total volume is \(100 \ ext{mL} + 100 \ ext{mL} = 100 \ ext{mL} \). The concentration of excess HCl is: \[ \text{Concentration} = \frac{0.001 \ ext{mol}}{0.1 \ ext{L}} = 0.01 \ ext{M} \]The pH is \(\text{pH} = -\log[H^+] = 2\).

Step 8: Calculate Moles for Option (c)

In option (c), the amount of HCl is far less than NaOH. HCl moles (0.001) are neutralized by an excess of NaOH moles (0.009). The solution will remain basic after neutralization with pH greater than 7.

Step 9: Calculate Moles for Option (d)

In option (d), the concentrations for NaOH and HCl are different. Calculate moles: - HCl: \[75 \ ext{mL} \times \frac{1 \text{L}}{1000 \text{mL}} \times 0.1 \text{M} = 0.0075 \text{mol} \]- NaOH: \[25 \ ext{mL} \times \frac{1 \text{L}}{1000 \text{mL}} \times 0.2 \text{M} = 0.005 \text{mol} \]

Step 10: Determine Excess HCl in Option (d)

In option (d), the moles of HCl exceed those of NaOH, hence there is an excess of \(0.0075 - 0.005 = 0.0025\ \text{mol}\) of HCl left.

Step 11: Calculate pH for Option (d)

The pH of the resulting solution in option (d) is determined by the concentration of excess HCl. The total volume is \(75 \ ext{mL} + 25 \ ext{mL} = 100 \ ext{mL} \).The concentration of excess HCl is: \[ \text{Concentration} = \frac{0.0025 \ ext{mol}}{0.1 \ ext{L}} = 0.025 \ ext{M} \]The pH is \(\text{pH} = -\log[H^+] \approx 1.6\).

Step 12: Determination of Closest pH

Option (d) yields a pH closest to 1. By calculating the excess moles and considering dissociation into hydrogen ions, option (d) with excess HCl results in a strong acid solution near pH 1.

Key Concepts

Acid-Base Neutralization

In solution chemistry, acid-base neutralization is a vital concept. It occurs when an acid reacts with a base to produce water and a salt. This process is essential for many chemical reactions and real-life applications.
Neutralization occurs because acids donate protons (\(H^+\)), while bases accept protons. For example, when hydrochloric acid (\(HCl\)) reacts with sodium hydroxide (\(NaOH\)), it forms water (\(H_2O\)) and sodium chloride (\(NaCl\)).
Neutralization is complete when the number of moles of acid equals the number of moles of base:

• If moles of acid = moles of base, the resulting solution is neutral with a pH of 7.
• If more acid is present, the solution is acidic, resulting in a pH less than 7.
• If more base is present, the solution is basic, resulting in a pH greater than 7.

This concept helps in calculations involving \(pH\) levels and determining whether a solution will be acidic or basic.

Molar Calculations

Molar calculations are fundamental in chemistry for determining the amount of substance involved in reactions. Moles are a unit that measures the amount of a chemical substance. The \(mole\) concept allows chemists to use the quantity of matter in calculations effectively.
To calculate moles:

• Use the formula \( ext{Moles} = rac{ ext{Volume (L)} imes ext{Molarity (M)}}{ ext{1 L}}\).
• Ensure volumes are converted into liters by dividing milliliters by 1000.
• Multiply the volume by the molarity to find the number of moles.

These calculations help in predicting reaction outcomes and balancing equations. For instance, in neutralization reactions, molar calculations determine if an acid completely reacts with a base.
Understanding molar calculations is vital for determining quantities and predicting the nature of produced solutions.

Solution Chemistry

Solution chemistry explores how substances dissolve and interact in a solvent, resulting in a solution. It’s a key part of understanding reactions in chemistry.
Solutions consist of a solute (substance) dissolved in a solvent (often water). The concentration of a solution is an essential concept and can be described in molarity (\(M\)), representing moles of solute per liter of solution.
Key features of solutions include:

• Solutions can be described as acidic, neutral, or basic based on their \(pH\).
• In aqueous solutions, water is the solvent.
• Concentrated solutions hold more solute, while dilute ones hold less.

The \(pH\) of a solution is crucial, showing its acidity or alkalinity. The \(pH\) scale ranges from 0 to 14:

• Acidic solutions have \(pH\) values less than 7.
• Neutral solutions have a \(pH\) equal to 7.
• Basic solutions have \(pH\) values greater than 7.

Learning about solution chemistry is essential in understanding how substances interact in chemical processes.