The total molarity and normality of all the ions present in a solution containing 0.1 M of `CuSO_4` and 0.1 M of `Al_2(SO_4)_3` is

Options

A. 0.2 M, 0.2 N
B. 0.7 M , 1.6 N
C. 0.8 M, 1.6 N
D. 1.6 M , 1.6 N

Answer

To find the total molarity and normality of all the ions present in a solution containing 0.1 M of CuSO₄ and 0.1 M of Al₂(SO₄)₃, we will follow these steps:

### Step 1: Determine the dissociation of CuSO₄
CuSO₄ dissociates in water as follows:
\[ \text{CuSO}_4 \rightarrow \text{Cu}^{2+} + \text{SO}_4^{2-} \]
- From 0.1 M of CuSO₄, we get:
- Molarity of Cu²⁺ = 0.1 M
- Molarity of SO₄²⁻ = 0.1 M

### Step 2: Determine the dissociation of Al₂(SO₄)₃
Al₂(SO₄)₃ dissociates in water as follows:
\[ \text{Al}_2(\text{SO}_4)_3 \rightarrow 2 \text{Al}^{3+} + 3 \text{SO}_4^{2-} \]
- From 0.1 M of Al₂(SO₄)₃, we get:
- Molarity of Al³⁺ = 2 × 0.1 M = 0.2 M
- Molarity of SO₄²⁻ = 3 × 0.1 M = 0.3 M

### Step 3: Calculate the total molarity of all ions
Now, we can sum the molarities of all ions:
- Molarity of Cu²⁺ = 0.1 M
- Molarity of SO₄²⁻ from CuSO₄ = 0.1 M
- Molarity of Al³⁺ = 0.2 M
- Molarity of SO₄²⁻ from Al₂(SO₄)₃ = 0.3 M

Total molarity of ions:
\[ \text{Total Molarity} = 0.1 + 0.1 + 0.2 + 0.3 = 0.7 \, \text{M} \]

### Step 4: Calculate the normality of each ion
Normality (N) is calculated using the formula:
\[ \text{Normality} = \text{Molarity} \times n \]
where n is the number of equivalents.

1. For Cu²⁺:
- Molarity = 0.1 M, n = 2 (charge)
- Normality = 0.1 × 2 = 0.2 N

2. For SO₄²⁻ from CuSO₄:
- Molarity = 0.1 M, n = 2 (charge)
- Normality = 0.1 × 2 = 0.2 N

3. For Al³⁺:
- Molarity = 0.2 M, n = 3 (charge)
- Normality = 0.2 × 3 = 0.6 N

4. For SO₄²⁻ from Al₂(SO₄)₃:
- Molarity = 0.3 M, n = 2 (charge)
- Normality = 0.3 × 2 = 0.6 N

### Step 5: Calculate the total normality
Now, we can sum the normalities of all ions:
- Normality of Cu²⁺ = 0.2 N
- Normality of SO₄²⁻ from CuSO₄ = 0.2 N
- Normality of Al³⁺ = 0.6 N
- Normality of SO₄²⁻ from Al₂(SO₄)₃ = 0.6 N

Total normality:
\[ \text{Total Normality} = 0.2 + 0.2 + 0.6 + 0.6 = 1.6 \, \text{N} \]

### Final Answer:
- Total Molarity = 0.7 M
- Total Normality = 1.6 N