Consider the following statements.
I. The pH of a mixture containing 400 mL of 0.1 M H2SO4 and 400 m

Q: Why did we multiply by 2 for $H_2SO_4$?

Sulfuric acid ($H_2SO_4$) is a strong dibasic acid. It completely dissociates to give two $H^+$ ions for every one molecule of the acid. Therefore, its normality (or $H^+$ concentration) is twice its molarity.

Q: Why does $K_w$ increase with temperature?

The auto-ionization of water is an endothermic process because energy is required to break the O-H bonds. According to Le-Chatelier's principle, increasing the temperature favors the endothermic (forward) direction, producing more $H^+$ and $OH^-$ ions, thus increasing $K_w$.

Q: Why couldn't we use the approximation $K_a = C\alpha^2$ for statement III?

The approximation $1 - \alpha \approx 1$ is only valid when $\alpha$ is very small (typically less than 5% or 0.05). In this problem, $\alpha$ turns out to be 50% (0.5), which is huge! Ignoring it in the denominator would lead to a completely wrong answer.

Q: What exactly is the common-ion effect?

It is the decrease in solubility or dissociation of a weak electrolyte when a strong electrolyte containing an ion in common with the weak electrolyte is added to the solution. It is a direct consequence of Le-Chatelier's principle.

Q: How did we calculate $\log 20$ without a calculator?

You can break it down using log properties: $\log 20 = \log(2 \times 10) = \log 2 + \log 10$. Since $\log 10 = 1$ and $\log 2 \approx 0.301$, the sum is $1.301$.

Question

Consider the following statements.
I. The pH of a mixture containing 400 mL of 0.1 M $H_2SO_4$ and 400 mL of 0.1 M NaOH will be approximately 1.3.
II. Ionic product of water is temperature dependent.
III. A monobasic acid with $K_a = 10^{-5}$ has a pH = 5. The degree of dissociation of this acid is 50%.
IV. The Le-Chatelier's principle is not applicable to common-ion effect.
The correct statements are

Options:

(A) I, II and IV

(B) II and III

(C) I and II

(D) I, II and III

Sigma Percentile · Accepted Answer

This question is a fantastic conceptual marathon that tests your grip on multiple facets of ionic equilibrium. Instead of just solving one numerical, we are tasked with evaluating four distinct statements. Let's break them down one by one and uncover the chemistry behind each.

### The Mixture's pH

Statement I asks us to find the pH of a mixture containing $400 	ext{ mL}$ of $0.1 	ext{ M } H_2SO_4$ and $400 	ext{ mL}$ of $0.1 	ext{ M NaOH}$.

The first thing to notice is that sulfuric acid ($H_2SO_4$) is a **dibasic acid**. This means each mole of $H_2SO_4$ provides two moles of $H^+$ ions. Let's calculate the millimoles of $H^+$ and $OH^-$ available for neutralization:

$$n_{H^+} = 400 	ext{ mL} 	imes 0.1 	ext{ M} 	imes 2 = 80 	ext{ mmol}$$

$$n_{OH^-} = 400 	ext{ mL} 	imes 0.1 	ext{ M} 	imes 1 = 40 	ext{ mmol}$$

Clearly, $H^+$ is in excess, making $OH^-$ the limiting reagent. After the neutralization reaction, the remaining millimoles of $H^+$ will be $80 - 40 = 40 	ext{ mmol}$.

The total volume of the mixture is now $400 + 400 = 800 	ext{ mL}$. We can find the final concentration of $H^+$:

$$[H^+] = \frac{40 	ext{ mmol}}{800 	ext{ mL}} = \frac{1}{20} 	ext{ M}$$

Now, we calculate the pH:

$$	ext{pH} = -\log\left(\frac{1}{20}ight) = \log 20$$

Using logarithm properties, $\log 20 = \log(2 	imes 10) = \log 2 + \log 10 = 0.301 + 1 = 1.301$.

This matches the statement perfectly. **Statement I is True.**

### The Heat of Water

Statement II claims that the ionic product of water ($K_w$) is temperature dependent.

Think about the auto-ionization of water:

$$H_2O(l) ightleftharpoons H^+(aq) + OH^-(aq)$$

Breaking the covalent bonds in water to form ions requires energy, making this an **endothermic process** ($\Delta H > 0$). According to **Le-Chatelier's principle**, if we increase the temperature of an endothermic reaction, the equilibrium shifts in the forward direction to absorb the excess heat.

As the reaction shifts forward, the concentrations of $H^+$ and $OH^-$ increase, which directly increases the value of $K_w = [H^+][OH^-]$. Therefore, $K_w$ is indeed highly dependent on temperature. **Statement II is True.**

### The Weak Acid's Secret

Statement III presents a weak monobasic acid with $K_a = 10^{-5}$ and a pH of $5$. We need to verify if its degree of dissociation ($\alpha$) is $50\%$.

A pH of $5$ immediately tells us that $[H^+] = 10^{-5} 	ext{ M}$. For a weak acid $HA$, the concentration of $H^+$ is given by $C\alpha$. So, $C\alpha = 10^{-5}$.

The exact expression for the acid dissociation constant is:

$$K_a = \frac{C\alpha^2}{1-\alpha}$$

We can cleverly rewrite this to use our known value of $C\alpha$:

$$K_a = \frac{(C\alpha) \cdot \alpha}{1-\alpha}$$

Substitute $K_a = 10^{-5}$ and $C\alpha = 10^{-5}$ into the equation:

$$10^{-5} = \frac{10^{-5} \cdot \alpha}{1-\alpha}$$

The $10^{-5}$ terms cancel out beautifully, leaving us with:

$$1 = \frac{\alpha}{1-\alpha}$$

Cross-multiplying gives $1 - \alpha = \alpha$, which simplifies to $2\alpha = 1$, or $\alpha = 0.5$. Converting this to a percentage, we get exactly $50\%$. **Statement III is True.**

### The Common Ion Fallacy

Statement IV asserts that Le-Chatelier's principle is not applicable to the common-ion effect.

This is fundamentally incorrect. The **common-ion effect** is defined as the suppression of the degree of dissociation of a weak electrolyte when a strong electrolyte containing a common ion is added.

For example, if you add $CH_3COONa$ to a solution of $CH_3COOH$, the high concentration of acetate ions ($CH_3COO^-$) from the salt acts as an added product. According to **Le-Chatelier's principle**, adding a product forces the equilibrium to shift backward to consume it, thereby suppressing the dissociation of the weak acid. The common-ion effect is literally a direct application of Le-Chatelier's principle! **Statement IV is False.**

Since statements I, II, and III are correct, the right choice is **Option (d)**.

Select Answer:

Visualized Solution

$Analyzing the Statements$

$Statement I: Millimoles Calculation$

$Statement I: pH of Mixture$

$Statement II: Ionic Product of Water$

$Statement III: Weak Acid Dissociation$

$Statement III: Degree of Dissociation$

$Statement IV: Le-Chatelier’s Principle$

$Final Answer$

The Sigma Insight: pH, Buffer and Indicator

The Mixture's pH

The Heat of Water

The Weak Acid's Secret

The Common Ion Fallacy

Similar Questions

For the following Assertion and Reason, the correct option is Assertion (A) The pH of water increases with increase in temperature. Reason (R) The dissociation of water into $H^{+}$ and $OH^{-}$ is an exothermic reaction.

Comprehension Passage

Enthalpy of dissociation (in $kJ mol^{- 1}$ ) of acetic acid obtained from the Expt-2 is

The $pH$ of the solution after Expt-2

Which one of the following statements is not true?

20 mL of 0.1 M $H_{2} SO_{4}$ solution is added to 30 mL of 0.2 M $NH_{4} OH$ solution. The pH of the resultant mixture is [ $pK_{b}$ of $NH_{4} OH = 4.7$ ]

The $p K_{a}$ of a weak acid (HA) is $4.5$ . The pOH of an aqueous buffered solution of HA in which $50%$ of the acid ionised is

Two solutions, $A$ and $B$ , each of $100 L$ was made by dissolving $4 g$ of $NaOH$ and $9.8 g$ of $H_{2} SO_{4}$ in water, respectively. The pH of the resultant solution obtained from mixing $40 L$ of solution $A$ and $10 L$ of solution $B$ is ……… .

In an acid-base titration, 0.1 M HCl solution was added to the NaOH solution of unknown strength. Which of the following correctly shows the change of pH of the titration mixture in this experiment?

100 mL of 0.1M HCl is taken is a beaker and to it 100 mL of 0.1 M NaOH is added in steps of 2 mL and the pH continuously measured. Which of the following graphs correctly depicts the change in pH?

An aqueous solution of 1M NaCl and 1M HCl is

How many litres of water must be added to 1 L of an aqueous solution of HCl with a pH of 1 to create an aqueous solution with pH of 2?

Select Answer:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Analyzing the Statements

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Statement I: Millimoles Calculation

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Statement I: pH of Mixture

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Statement II: Ionic Product of Water

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Statement III: Weak Acid Dissociation

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Statement III: Degree of Dissociation

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Statement IV: Le-Chatelier’s Principle

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Final Answer