Problem 59 - 43 Years JEE ADVANCED (1978-2020) + JEE MAIN Chapterwise & Topicwise Solved Papers Chemistry

Question

What will be the resultant \(\mathrm{pH}\) when \(200 \mathrm{~mL}\) of an aqueous solution of \(\mathrm{HCl}(\mathrm{pH}=2.0)\) is mixed with \(300 \mathrm{~mL}\) of an aqueous solution of \(\mathrm{NaOH}(\mathrm{pH}=12.0) ?\)

Step-by-Step Solution

Answer

The resultant pH of the mixture is approximately 11.3.

Step 1: Calculate Moles of HCl

First, determine the concentration of \(\text{HCl}\) using its \(\text{pH}\). We have: \(pH = 2.0\), hence the \([ ext{H}^+] = 10^{-2} \text{ M}\). The moles of \(\text{HCl}\) are calculated by multiplying its concentration by its volume. Volume in liters (L): \(200 \, \text{mL} = 0.200 \, \text{L}\), so the moles of \(\text{HCl}\) are \((10^{-2} \, \text{mol/L}) \, \times \, (0.200 \, \text{L}) = 2 \, \times \, 10^{-3} \, \text{mol}\).

Step 2: Calculate Moles of NaOH

Now, perform a similar calculation for \(\text{NaOH}\). Given its \(pH = 12.0\),we can find \(\text{pOH} = 14 - 12 = 2\) and \([ ext{OH}^-] = 10^{-2} \, \text{M}\). Volume in liters is \(300 \, \text{mL} = 0.300 \, \text{L}\), hence the moles of \(\text{NaOH}\) are \((10^{-2} \, \text{mol/L}) \, \times \, (0.300 \, \text{L}) = 3 \, \times \, 10^{-3} \, \text{mol}\).

Step 3: Determine Limiting Reactant

To find the limiting reactant, compare moles of \(\text{HCl}\) and \(\text{NaOH}\).Both react in a 1:1 ratio, thus \(2 \, \times \, 10^{-3}\text{ mol} \text{HCl}\) and \(3 \, \times \, 10^{-3}\text{ mol} \text{NaOH}\).Since \(\text{HCl}\) is less, it is the limiting reactant.

Step 4: Calculate Excess NaOH

Calculate the remaining \(\text{NaOH}\) after neutralization.The excess \(\text{NaOH}\) moles = \((3 \, \times \, 10^{-3} - 2 \, \times \, 10^{-3}) = 1 \, \times \, 10^{-3} \, \text{mol}\).

Step 5: Calculate New Concentration of OH\(^{-}\)

Find the concentration of remaining \(\text{OH}^-\) in the combined solution.The total volume of the mixture is \((200 + 300) \, \text{mL} = 500 \, \text{mL} = 0.500 \, \text{L}\).So, the concentration of \(\text{OH}^-\) is \((1 \, \times \, 10^{-3} \, \text{mol}) / (0.500 \, \text{L}) = 2 \, \times \, 10^{-3} \, \text{M} \).

Step 6: Calculate Resultant pH

Convert \([ ext{OH}^-]\) to \(\text{pOH}\): \(pOH = -\log(2 \, \times \, 10^{-3}) \approx 2.70\).Find the \(\text{pH}\) using \(\text{pH} + ext{pOH} = 14\), so \(\text{pH} = 14 - 2.70 \approx 11.3\).

Key Concepts

Acid-Base Neutralization

Acid-base neutralization is a fundamental chemical process where an acid and a base react to form water and a salt. This process typically involves the reaction of hydrogen ions \(\text{(H}^+\text{)}\) from the acid and hydroxide ions \(\text{(OH}^-\text{)}\) from the base. When these two types of ions come together, they form water, \(\text{H}_2\text{O}\). In the exercise given, we have hydrochloric acid (HCl) as the acid and sodium hydroxide (NaOH) as the base. When these two react in a 1:1 ratio, they neutralize each other:

• HCl donates \(\text{H}^+\text{ ions}\).
• NaOH provides \(\text{OH}^-\text{ ions}\).

The equation is:\[\text{HCl} + \text{NaOH} \rightarrow \text{NaCl} + \text{H}_2\text{O}\]Both solutions are mixed until one component, the limiting reactant, is consumed.

Limiting Reactant

The limiting reactant in a chemical reaction is the substance that is completely used up first, thus limiting the amount of product that can be formed. Identifying the limiting reactant is crucial because it determines the extent of the chemical reaction. Here is how we identify it:

• Calculate the moles of each reactant.
• Compare them based on the stoichiometry of the reaction.

For the problem at hand, after calculating the moles, we find that HCl has fewer moles (2 x 10⁻³) than NaOH (3 x 10⁻³). Therefore, HCl is the limiting reactant because there is less of it available to react with NaOH completely. The result is that some NaOH remains unreacted after the HCl is used up. Determining which substance runs out first helps predict how much of the other reactants or products will still be present at the end of the reaction.

Moles Calculation

Calculating the number of moles is an essential part of solving chemical reaction problems. A mole is a unit that chemists use to describe amounts of a substance, typically reflecting its number of atoms, molecules, or ions. The relationship between volume, concentration, and moles is given by:\[\text{Moles} = \text{Concentration} \times \text{Volume}\]Here, we begin by finding the concentration of each solution from the given pH:

• HCl has a \(\text{H}^+\text{ concentration of }10^{-2} \text{ M}\) because of its pH of 2.
• NaOH has an \(\text{OH}^-\text{ concentration of }10^{-2} \text{ M}\) deduced from a pH of 12 (pOH = 2).

We can then use the known volumes of 200 mL for HCl and 300 mL for NaOH to find moles:

• Moles of HCl: \(0.200 \text{ L} \times 10^{-2} \text{ M} = 2 \times 10^{-3} \text{ mol}\)
• Moles of NaOH: \(0.300 \text{ L} \times 10^{-2} \text{ M} = 3 \times 10^{-3} \text{ mol}\)

Concentration Determination

Concentration determination is the calculation of the amount of solute present per unit volume of solution. After the reaction, determining the concentration of any remaining reactants or products is crucial for further calculations, such as pH.Once the limiting reactant (HCl) is used up, we calculate the concentration of the remaining NaOH. Here's what we need to do:

• Find the total volume of the combined solution, here 500 mL or 0.500 L.
• Determine remaining moles of NaOH, which was not completely neutralized (1 x 10⁻³ moles).

The concentration is calculated by:\[\text{Concentration of remaining } \text{OH}^- = \frac{1 \times 10^{-3} \text{ mol}}{0.500 \text{ L}} = 2 \times 10^{-3} \text{ M}\]This concentration is used to find the pH of the solution, considering excess \(\text{OH}^-\text{ ions}\). The formula \(\text{pH} + \text{pOH} = 14\) helps convert \(\text{OH}^-\) concentration to \(\text{pH}\) using additional logarithmic calculations.