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 find the molarity of the hypo (sodium thiosulfate) solution used in the iodometric estimation, we will follow these steps:

### Step 1: Write down the reactions
The two reactions involved in the iodometric estimation 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 \( Cu^{2+} \)
From the question, we know that 0.12 moles of \( CuSO_4 \) were added. Since \( CuSO_4 \) dissociates to give \( Cu^{2+} \), we have:
- Moles of \( Cu^{2+} = 0.12 \, \text{moles} \)

### Step 3: Calculate moles of \( I_2 \) produced
According to the first reaction, 2 moles of \( Cu^{2+} \) produce 1 mole of \( I_2 \):
\[ \text{Moles of } I_2 = \frac{0.12 \, \text{moles of } Cu^{2+}}{2} = 0.06 \, \text{moles of } I_2 \]

### Step 4: Relate moles of \( I_2 \) to moles of \( Na_2S_2O_3 \)
From the second reaction, 1 mole of \( I_2 \) reacts with 2 moles of \( Na_2S_2O_3 \):
\[ \text{Moles of } Na_2S_2O_3 = 2 \times \text{Moles of } I_2 = 2 \times 0.06 = 0.12 \, \text{moles} \]

### Step 5: Calculate the molarity of the hypo solution
We know that the volume of the hypo solution used is 120 mL, which is equivalent to 0.120 L. Molarity (M) is calculated using the formula:
\[ \text{Molarity} = \frac{\text{moles of solute}}{\text{volume of solution in liters}} \]
Substituting the values:
\[ \text{Molarity of } Na_2S_2O_3 = \frac{0.12 \, \text{moles}}{0.120 \, \text{L}} = 1.0 \, \text{M} \]

### Final Answer
The molarity of the hypo solution is **1.0 M**.

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