Problem 256 - A Complete Resource Book in Chemistry for JEE Main

Question

What is the \(\mathrm{pH}\) of a mixture obtained by mixing \(75 \mathrm{ml}\) \(\mathrm{M} / 5 \mathrm{HCl}\) and \(25 \mathrm{ml} \mathrm{M} / 5 \mathrm{NaOH}\) aqueous solution?

Step-by-Step Solution

Answer

The pH of the mixture is 1.

Step 1: Calculate Moles of HCl

First, calculate the number of moles of HCl in the solution. The concentration of the HCl is \( \frac{1}{5} \text{ M} \) which is the same as \( 0.2 \text{ M} \). The volume of the HCl solution is \( 75 \text{ ml} \), which is \( 0.075 \text{ L} \) (since 1000 ml = 1 L). Calculate the moles as: \[ \text{moles of HCl} = 0.2 \text{ M} \times 0.075 \text{ L} = 0.015 \text{ moles} \].

Step 2: Calculate Moles of NaOH

Similarly, calculate the moles of NaOH. The concentration of NaOH is also \( \frac{1}{5} \text{ M} \) or \( 0.2 \text{ M} \), and the volume of the NaOH solution is \( 25 \text{ ml} \), which is \( 0.025 \text{ L} \). Calculate the moles as: \[ \text{moles of NaOH} = 0.2 \text{ M} \times 0.025 \text{ L} = 0.005 \text{ moles} \].

Step 3: Determine the Remaining HCl or NaOH

HCl and NaOH react according to the equation \( \text{HCl} + \text{NaOH} \rightarrow \text{NaCl} + \text{H}_2\text{O} \). Using the mole numbers calculated:- HCl: 0.015 moles- NaOH: 0.005 molesAfter reaction, the excess HCl remaining is: \[ 0.015 \text{ moles} - 0.005 \text{ moles} = 0.01 \text{ moles} \].

Step 4: Determine Final Volume and Concentration

The total volume of the mixture is the sum of the volumes of HCl and NaOH: \( 75 \text{ ml} + 25 \text{ ml} = 100 \text{ ml} \) or \( 0.1 \text{ L} \). Thus, the concentration of the remaining HCl is: \[ \text{Concentration of HCl} = \frac{0.01 \text{ moles}}{0.1 \text{ L}} = 0.1 \text{ M} \].

Step 5: Calculate the pH

The pH is calculated using the formula \( \text{pH} = -\log[\text{H}^+] \). For the HCl solution, \([\text{H}^+]\) is the same as the concentration of HCl, which is \( 0.1 \text{ M} \). Thus, the pH is: \[ \text{pH} = -\log(0.1) = 1 \].

Key Concepts

Acid-Base Reaction

An acid-base reaction is an important type of chemical reaction where an acid reacts with a base to form salt and water. This process is commonly referred to as neutralization. In the exercise, hydrochloric acid (HCl) and sodium hydroxide (NaOH) undergo an acid-base reaction.

• HCl is a strong acid, meaning it completely dissociates in water to give hydrogen ions (H^+).
• NaOH is a strong base, providing hydroxide ions (OH^-) in the solution.

The reaction can be written as:\[\text{HCl}_{(aq)} + \text{NaOH}_{(aq)} \rightarrow \text{NaCl}_{(aq)} + \text{H}_2\text{O}_{(l)}\]Every HCl molecule reacts with NaOH to form one molecule each of sodium chloride (NaCl), a neutral salt, and water (H}_2\text{O). The key point here is that these types of reactions generally produce water, a non-reactive outcome in terms of pH calculation, mainly focusing the change on the remaining excess substance after neutralization takes place.

Mole Calculation

Mole calculation is crucial in chemistry as it allows us to quantify how much of a substance is involved in a reaction. The mole is a basic unit that describes the amount of substance, connecting atoms and molecules to macroscopic amounts. In the problem, calculating the moles of HCl and NaOH helps determine the extent of the reaction.Steps to calculate moles:

• Use the formula: \( \text{moles} = \text{Concentration (M)} \times \text{Volume (L)}\)
• Convert volume from milliliters to liters (1 L = 1000 ml).
• For HCl: \[0.2 \text{ M} \times 0.075 \text{ L} = 0.015 \text{ moles} \]
• For NaOH: \[0.2 \text{ M} \times 0.025 \text{ L} = 0.005 \text{ moles}\]

This sets the stage to find the remaining moles, which indicates the completion of the neutralization process.

Stoichiometry

Stoichiometry plays a central role in balancing chemical equations, which aids in predicting the products of a reaction and quantities needed. In acid-base reactions, stoichiometry allows us to align theoretical predictions with practical observations. Key aspects of stoichiometry:

• It relates the quantities of reactants and products.
• Based on the balanced equation: \( \text{HCl} + \text{NaOH} \rightarrow \text{NaCl} + \text{H}_2\text{O} \), it’s evident the reaction ratio is 1:1.
• According to stoichiometry, 0.005 moles of NaOH reacts with an equivalent 0.005 moles of HCl, leaving an excess of 0.01 moles of HCl (since initial HCl was 0.015 moles).

This stoichiometric balance leads to simple arithmetic to determine which reactant remains and its subsequent pH influence.

Hydrochloric Acid

Hydrochloric acid ( HCl ) is a common strong acid found in both laboratory and industrial applications. Its strength is derived from its complete ionization in water. When discussing pH and reactions, HCl is a crucial point of focus.

• HCl completely dissociates in water, meaning for every one mole of HCl , there is one mole of hydrogen ions ( H^+ ) produced.
• The concentration of H^+ ions determines the pH of the solution. In this exercise, the pH calculation stems from knowing the concentration of remaining HCl .
• In this problem's reaction with NaOH , 0.015 moles of HCl initially engaged, with 0.01 moles of it remaining after achieving neutralization.

In this scenario, it's the leftover HCl that directly influences the final pH value of the solution.

Sodium Hydroxide

Sodium hydroxide ( NaOH ) is a typical strong base used across chemistry. Like HCl , NaOH also completely dissociates when dissolved in water, providing hydroxide ions ( OH^- ). Characteristics of NaOH :

• NaOH completely ionizes to produce one mole of OH^- ions per mole of NaOH .
• In the given reaction, NaOH contributes 0.005 moles of hydroxide ions to react with the equivalent moles of hydrogen ions from HCl .
• Once neutralization occurs, there are no leftover NaOH molecules, leaving us only to consider the remaining HCl for pH determination.

Understanding this basic helps solidify your grasp on reaction dynamics and outcomes when strong acids and bases meet.

Solution Chemistry

Solution chemistry involves understanding how substances dissolve and react in solvents. It heavily focuses on concentration, pH, and dissociation of substances. In the context of pH calculation, knowing how different components in a solution interact is essential. Key elements about solutions:

• Concentration is expressed in molarity ( M ), which stands for moles of solute per liter of solution.
• The mixture's final volume is 0.1 L , coming from adding 75 ml of HCl and 25 ml of NaOH .
• Understanding what remains in solution post-reaction (excess HCl ) is necessary for calculating the solution's pH.

Being proficient in reading and manipulating these numbers can guide you through complex reaction problems, ensuring accurate predictions of solution behaviors.