Problem 91 - 43 Years JEE ADVANCED (1978-2020) + JEE MAIN Chapterwise & Topicwise Solved Papers Chemistry

Question

Twenty \(\mathrm{mL}\) of \(0.2 \mathrm{M}\) sodium hydroxide is added to \(50 \mathrm{~mL}\) of \(0.2 \mathrm{M}\) acetic acid to give \(70 \mathrm{~mL}\) of the solution. What is the \(\mathrm{pH}\) of this solution? Calculate the additional volume of \(0.2 \mathrm{M} \mathrm{NaOH}\) required to make the \(\mathrm{pH}\) of the solution 4.74. The ionization constant of acetic acid is \(1.8 \times 10^{-5}\).

Step-by-Step Solution

Answer

The solution needs an additional 5 mL of 0.2 M NaOH to reach a pH of 4.74.

Step 1: Calculate Initial Moles of NaOH and Acetic Acid

First, calculate the moles of sodium hydroxide (NaOH) and acetic acid (CH₃COOH) present initially. Use the formula: moles = concentration \(\times\) volume (in liters).For NaOH: \(0.2 \text{ M} \times 0.020 \text{ L} = 0.004 \text{ moles}\).For CH₃COOH: \(0.2 \text{ M} \times 0.050 \text{ L} = 0.010 \text{ moles}\).

Step 2: Determine Moles After Reaction

NaOH will react with CH₃COOH according to the equation: \(\text{NaOH} + \text{CH}_3\text{COOH} \rightarrow \text{CH}_3\text{COONa} + \text{H}_2\text{O}\).Since \(0.004\) moles of NaOH react completely with \(0.004\) moles of CH₃COOH, we are left with: Residual CH₃COOH: \(0.010 - 0.004 = 0.006 \text{ moles}\).Formed CH₃COONa (the salt): \(0.004 \text{ moles}\).

Step 3: Calculate Initial pH of the Solution

The solution now contains 0.006 moles of CH₃COOH and 0.004 moles of CH₃COONa in a total volume of 0.070 L.Use the Henderson-Hasselbalch equation: \[pH = pK_a + \log\left(\frac{[\text{A}^-]}{[\text{HA}] }\right)\]Where:\(pK_a = -\log(1.8 \times 10^{-5}) = 4.74\)Concentration of CH₃COONa, \([\text{A}^-] \)= \(\frac{0.004}{0.07}\approx 0.057 \text{ M}\)Concentration of CH₃COOH, \([\text{HA}]\) = \(\frac{0.006}{0.07} \approx 0.086 \text{ M}\)Calculate pH: \[pH = 4.74 + \log\left(\frac{0.057}{0.086}\right) = 4.511\]

Step 4: Determine Additional Volume for Desired pH

The desired pH is 4.74, which is the pKa, meaning [\(\text{A}^-\)] = [\(\text{HA}\)] for the Henderson-Hasselbalch equation: \[pH = pK_a + \log\left(\frac{[\text{A}^-]}{[\text{HA}] }\right) = 4.74\]This implies: \[\frac{[\text{A}^-]}{[\text{HA}] } = 1\]Thus, moles of CH₃COOH left should equal moles of CH₃COONa formed (new equilibrium state): Adding more NaOH will convert some CH₃COOH to CH₃COONa.Let additional moles of NaOH added be \(x\), then: \(0.006 - x = 0.004 + x\) which solves to \(x = 0.001\) moles extra NaOH needed. Convert to volume using 0.2M NaOH: \[x = 0.001 \text{ moles} \rightarrow 0.001 \div 0.2 = 0.005 \text{ L} = 5 \text{ mL}\]Add 5 mL of 0.2M NaOH.

Key Concepts

Henderson-Hasselbalch equation

The Henderson-Hasselbalch equation is a fundamental equation used in chemistry to estimate the pH of a buffer solution. A buffer solution is a special solution that resists changes in pH when small amounts of acid or base are added. The equation is expressed as: \[ pH = pK_a + \log\left(\frac{[\text{A}^-]}{[\text{HA}] }\right) \]Here, \( pH \) represents the potential of hydrogen, which tells us how acidic or basic the solution is. \( pK_a \) is the negative logarithm of the acid dissociation constant (\( K_a \)), providing a measure of the strength of an acid in solution. \([\text{A}^-]\) is the concentration of the conjugate base, and \([\text{HA}]\) is the concentration of the acid.
The Henderson-Hasselbalch equation is particularly useful in calculating the pH of solutions during acid-base reactions, especially when mixing weak acids with their conjugate bases. It simplifies complex equilibrium calculations by assuming that the concentrations of the acid and its conjugate base do not change significantly when the pH is near the \( pK_a \) value.

Buffer solution

A buffer solution is a type of chemical solution that helps maintain a stable pH level by neutralizing small quantities of added acid or base. This stability is achieved due to the presence of both a weak acid and its conjugate base (or a weak base and its conjugate acid). For example, a solution made from acetic acid (\(CH_3COOH\)) and sodium acetate (\(CH_3COONa\)) acts as a buffer.
Key characteristics of buffer solutions include:

• They resist pH change when small amounts of acid or base are added.
• They usually consist of a weak acid and its conjugate base or a weak base and its conjugate acid.
• Their effectiveness depends on the concentration and ratio of the acid to base species.

Buffer solutions are essential in many biological systems and chemical applications where a stable pH is required, such as in blood and many enzymatic reactions. The pH of a buffer can be effectively calculated using the Henderson-Hasselbalch equation, allowing predictions of how the system will react to changes.

Acid-base reaction

An acid-base reaction involves the transfer of hydrogen ions between reactants, transforming acids to bases and vice versa. These reactions are common in chemistry and biology. In an acid-base reaction, an acid donates a hydrogen ion \( (H^+) \) to a base, which accepts it. For example, acetic acid (\(CH_3COOH\)) reacts with sodium hydroxide (\(NaOH\)), transferring a \(H^+\) ion and forming water and the sodium acetate salt (\(CH_3COONa\)).
Key points to understand for acid-base reactions include:

• Strong acids or bases will completely ionize in water, while weak acids or bases only partially dissociate.
• The reaction between an acid and a base often results in the production of water and a salt.
• Acid-base reactions are governed by the principle of neutralization where acid and base neutralize each other's effect.

Understanding acid-base reactions is crucial for predicting the outcome of chemical processes, from laboratory chemistry to natural biological functions.

Ionization constant

The ionization constant, often denoted as \( K_a \), is a quantitative measure of the strength of an acid in solution. It represents the equilibrium constant for the dissociation of a weak acid into its constituent ions in a solution:\[ HA \rightleftharpoons H^+ + A^- \] Here, \( HA \) is a weak acid that ionizes into \(H^+\) and \(A^-\). The \( K_a \) value is determined by the concentrations of these ions at equilibrium:\[ K_a = \frac{[H^+][A^-]}{[HA]} \] The larger the \( K_a \), the stronger the acid, meaning it ionizes more in solution. A small \( K_a \) means the acid is weak and ionizes less, keeping most of its molecules intact.
The ionization constant is a crucial concept when dealing with acids in chemistry, as it helps chemists understand and predict how acids will behave in different conditions. Knowing the \( K_a \) and its relationship with its \( pK_a \) \((pK_a = -\log K_a)\) aids in calculating the pH of solutions using the Henderson-Hasselbalch equation.