In a iodomeric estimation, the following reactions occur
`2Cu^(2+)+4i^(-) to Cu_(2)I_(2)+I_(2), I_(2)+2Na_(2)S_(2)O_(3) to 2NaI+Na_(2)S_(4)O_(6)`
0.12 mole of `CuSO_(4)` was adde to excess of KI solution and the liberated iodine required 120mL of hypo. The molarity of hypo soulution was:

Options

A. 2
B. 0.2
C. 0.1
D. 1

Answer

To solve the problem, we will follow these steps:

### Step 1: Understand the Reactions
The reactions given are:
1. \( 2Cu^{2+} + 4I^{-} \rightarrow Cu_2I_2 + I_2 \)
2. \( I_2 + 2Na_2S_2O_3 \rightarrow 2NaI + Na_2S_4O_6 \)

### Step 2: Determine Moles of Copper Sulfate
We are given that \( 0.12 \) moles of \( CuSO_4 \) were used. This means that the moles of \( Cu^{2+} \) ions are also \( 0.12 \) moles.

### Step 3: Calculate the Number of Equivalents of Copper
The n-factor for \( Cu^{2+} \) is \( 2 \) because it changes from \( +2 \) to \( +1 \) (a change of 1 for each copper ion, and there are 2 copper ions in the reaction). Thus, the number of equivalents of \( Cu^{2+} \) is calculated as:
\[ \text{Number of equivalents of } Cu^{2+} = \text{Number of moles} \times n\text{-factor} = 0.12 \times 2 = 0.24 \text{ equivalents} \]

### Step 4: Calculate the Number of Moles of Iodine Liberated
From the first reaction, \( 2 \) equivalents of \( Cu^{2+} \) liberate \( 1 \) equivalent of \( I_2 \). Therefore, the number of equivalents of \( I_2 \) liberated is:
\[ \text{Number of equivalents of } I_2 = \frac{0.24}{2} = 0.12 \text{ equivalents} \]
Since \( 1 \) equivalent of \( I_2 \) corresponds to \( 1 \) mole of \( I_2 \), the moles of \( I_2 \) liberated is also \( 0.12 \).

### Step 5: Calculate the Number of Equivalents of Hypo Solution
The second reaction shows that \( 1 \) mole of \( I_2 \) reacts with \( 2 \) moles of \( Na_2S_2O_3 \). Therefore, the number of equivalents of \( Na_2S_2O_3 \) used is:
\[ \text{Number of equivalents of } Na_2S_2O_3 = 0.12 \times 2 = 0.24 \text{ equivalents} \]

### Step 6: Relate the Equivalents to Molarity
We know that:
\[ \text{Number of equivalents} = \text{Molarity} \times \text{Volume (L)} \]
Given that the volume of hypo solution used is \( 120 \, \text{mL} = 0.120 \, \text{L} \), we can set up the equation:
\[ 0.24 = M \times 0.120 \]

### Step 7: Solve for Molarity
Rearranging the equation gives:
\[ M = \frac{0.24}{0.120} = 2 \, \text{M} \]

### Final Answer
The molarity of the hypo solution is \( 2 \, \text{M} \).

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