Question

0.1 M solution of KI reacts with excess of \( \text{H}_2\text{SO}_4 \) and KIO\(_3\), according to the equation: \[ 5I^- + 6H^+ \rightarrow 3I_2 + 3H_2O \]

Identify the correct statements:

• (A) 200 mL KI solution reacts with 0.004 mol of KIO₃
• (B) 200 mL KI solution reacts with 0.006 mol of H₂SO₄
• (3) 0.5 L KI solution produced 0.005 mol of I₂
• (4) Equivalent weight of KIO₃ is equal to Molecular weight / 5

• (A) and (D) only
• (B) and (C) only
• (A) and (B) only
• (C) and (D) only

Hint:

The equivalent weight of a substance is defined as its molecular weight divided by the number of electrons involved in the reaction. For KIO\(_3\), 5 moles of electrons are involved per mole of KIO\(_3\).

Answer 1

The Correct Option is A

- Statement (A) is correct because 200 mL of 0.1 M KI contains 0.02 moles of KI, and according to the equation, 5 moles of I\(^-\) react with 1 mole of KIO\(_3\). Therefore, 0.02 moles of KI would require 0.004 mol of KIO\(_3\).
- Statement (C) is also correct because 0.5 L of 0.1 M KI will contain 0.05 moles of KI, and according to the equation, this will produce 0.005 mol of I\(_2\).
- Statement (D) is correct because the equivalent weight of KIO\(_3\) is equal to its molecular weight divided by 5, as 5 moles of iodide react with one mole of KIO\(_3\).
Thus, the correct answer is \( \boxed{(A),(D)} \).

Answer 2

Step 1: Understand the given reaction.
The balanced reaction is:
\[ 5I^- + IO_3^- + 6H^+ \rightarrow 3I_2 + 3H_2O \] Here, iodide ions (\( I^- \)) are oxidized to iodine (\( I_2 \)) and iodate ions (\( IO_3^- \)) are reduced. The acid (\( H_2SO_4 \)) provides \( H^+ \) ions required for the reaction.

Step 2: Given data.
Molarity of KI solution = 0.1 M
Volume of KI solution = 200 mL = 0.2 L
Moles of \( I^- \) in solution:
\[ n_{I^-} = 0.1 \times 0.2 = 0.02 \, \text{mol} \] From the balanced equation:
5 moles of \( I^- \) react with 1 mole of \( IO_3^- \).
Hence, moles of \( IO_3^- \) required = \( \frac{0.02}{5} = 0.004 \, \text{mol} \).
Therefore, 200 mL of 0.1 M KI reacts with 0.004 mol of KIO₃.
Thus, statement (A) is true.

Step 3: Check statement (B).
From the equation, 5 moles of \( I^- \) require 6 moles of \( H^+ \).
So, for 0.02 mol of \( I^- \), moles of \( H^+ \) needed are:
\[ \frac{6}{5} \times 0.02 = 0.024 \, \text{mol} \] Since each \( H_2SO_4 \) molecule provides 2 \( H^+ \) ions:
Moles of \( H_2SO_4 \) required = \( \frac{0.024}{2} = 0.012 \, \text{mol} \).
Statement (B) says 0.006 mol of \( H_2SO_4 \), which is incorrect.
Hence, statement (B) is false.

Step 4: Check statement (C).
From the equation, 5 mol of \( I^- \) produce 3 mol of \( I_2 \).
For 0.5 L of 0.1 M KI, moles of \( I^- = 0.05 \).
Moles of \( I_2 \) produced = \( \frac{3}{5} \times 0.05 = 0.03 \, \text{mol} \).
Statement (C) says 0.005 mol of \( I_2 \), which is incorrect.
Hence, statement (C) is false.

Step 5: Check statement (D).
In this reaction, 1 mole of \( IO_3^- \) reacts by gaining 5 electrons (since 5 \( I^- \) are oxidized).
Therefore, the equivalent weight of \( KIO_3 \) is:
\[ \text{Equivalent weight} = \frac{\text{Molecular weight}}{5} \] Hence, statement (D) is true.

Step 6: Conclusion.
The correct statements are (A) and (D) only.

Final Answer:
\[ \boxed{\text{(A) and (D) only}} \]