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 \( \mathrm{pH} \) of the mixture is 1.

Step 1: Calculate Moles of HCl

First, find the moles of hydrochloric acid (HCl) in the solution. The concentration of HCl is given as \( \frac{M}{5} \), which means \( M \) divided by 5. Assuming \( M = 1 \), its concentration is \( 0.2 \space \text{mol/L} \). The volume of the HCl solution is 75 mL, or 0.075 L. Calculate moles using \( \text{moles} = \text{concentration} \times \text{volume} \). Thus, \( \text{moles of HCl} = 0.2 \times 0.075 = 0.015 \).

Step 2: Calculate Moles of NaOH

Next, calculate the moles of sodium hydroxide (NaOH) in the solution. Similarly, the concentration of NaOH is \( \frac{M}{5} = 0.2 \space \text{mol/L} \). The volume of the NaOH solution is 25 mL, or 0.025 L. Calculate moles using \( \text{moles} = \text{concentration} \times \text{volume} \). Thus, \( \text{moles of NaOH} = 0.2 \times 0.025 = 0.005 \).

Step 3: Determine Excess Moles of HCl

Determine the excess moles of HCl by subtracting the moles of NaOH from the moles of HCl, because HCl and NaOH neutralize each other. The excess moles of HCl will be \( 0.015 - 0.005 = 0.01 \) moles.

Step 4: Calculate Total Volume of the Mixture

Find the total volume of the mixture by adding the volumes of the two solutions: \( 75 \space \text{mL} + 25 \space \text{mL} = 100 \space \text{mL} \), or \( 0.1 \space \text{L} \).

Step 5: Calculate Concentration of Excess HCl

Find the concentration of the excess HCl by dividing the number of moles by the total volume of the solution. The concentration is \( \frac{0.01}{0.1} = 0.1 \space \text{M} \).

Step 6: Determine mathrm{pH} from Concentration

Calculate the \( \mathrm{pH} \) of the solution using the formula \( \mathrm{pH} = -\log[\text{H}^+] \). Since the concentration of \( \text{H}^+ \) ions from undissociated HCl is \( 0.1 \space \text{M} \), the \( \mathrm{pH} \) is \( -\log(0.1) = 1 \).

Key Concepts

Acid-Base Neutralization

When an acid meets a base, they neutralize each other—a process known as acid-base neutralization. This is like when two opposite teams on a seesaw balance each other out. In a neutralization reaction, an acid and a base react to form water and a salt, effectively canceling each other’s acidic and basic properties.
For instance, in the given problem, we have an acid: hydrochloric acid (HCl), and a base: sodium hydroxide (NaOH). When mixed together, these react to form water ( H_2O) and sodium chloride ( NaCl), a neutral compound.

• The formula for a general acid-base reaction is: Acid + Base → Salt + Water.
• The exact product and neutrality depend on the amounts used.

In this scenario, HCl was present in a larger quantity than NaOH, leaving an excess of HCl, which is responsible for the overall acidity in the solution.

Moles Calculation

In chemistry, calculating moles is crucial for understanding the amounts of reactants and products in a reaction. A mole represents a quantity of chemical substance related to its atomic weight and is equivalent to Avogadro’s number, which is approximately 6.022 x 10²³ particles.
To calculate moles:

• Use the formula: Moles = Concentration (C) × Volume (V).
• Concentration is often in mol/L (molarity), and volume should be in liters.

In the problem, we calculated moles of HCl and NaOH separately since both have the same concentration (0.2 mol/L) but different volumes:

• For HCl: Moles = 0.2 mol/L × 0.075 L = 0.015 moles.
• For NaOH: Moles = 0.2 mol/L × 0.025 L = 0.005 moles.

By comparing these amounts, we determined which substance has excess moles left after neutralization, essential for calculating the final pH.

Concentration Calculation

Concentration gives us the measurement of how much of a substance is present within a given volume of solution. It's generally expressed in molarity (M), which is moles per liter.
After neutralization, we had some hydrochloric acid left over. The concentration of this excess acid determines the acidity, represented by its H^+ ion concentration in the solution.

• To calculate concentration, use: Concentration = Moles ÷ Total Volume.
• Remember to express volume in liters (L) for accurate results.

In our exercise, we calculated the excess moles of HCl (0.01 moles) against the total mixture volume of 0.1 L, which gave us 0.1 M for the concentration of excess HCl. This concentration value is then used to find the pH of the mixture, defining its acidic strength.