The solubility of `Ba(OH)_(2)`. `8H_(2)O` in water at `288K` is `5.6g` per `100g` of water. What is the molality hydroxide ions in saturated solution of `Ba(OH)_(2)`. `8H_(2)O` at `288K` ? [At. Mass of `Ba=137,O=16,H=1`)

Explanation

To find the molality of hydroxide ions in a saturated solution of `Ba(OH)2 · 8H2O` at `288 K`, we can follow these steps:

### Step 1: Calculate the molar mass of `Ba(OH)2 · 8H2O`
- The molar mass of `Ba(OH)2 · 8H2O` can be calculated as follows:
- Atomic mass of Ba = 137 g/mol
- Atomic mass of O = 16 g/mol
- Atomic mass of H = 1 g/mol
- Molar mass of `OH` = 16 + 1 = 17 g/mol
- Molar mass of `Ba(OH)2` = 137 + 2(17) = 137 + 34 = 171 g/mol
- Molar mass of `8H2O` = 8 × (2 + 16) = 8 × 18 = 144 g/mol
- Total molar mass of `Ba(OH)2 · 8H2O` = 171 + 144 = 315 g/mol

### Step 2: Calculate the molality of `Ba(OH)2 · 8H2O`
- Given that the solubility of `Ba(OH)2 · 8H2O` is 5.6 g per 100 g of water, we can calculate the molality using the formula:

\[ \text{Molality} (m) = \frac{\text{Number of moles of solute}}{\text{Mass of solvent in kg}} \]

- First, we find the number of moles of `Ba(OH)2 · 8H2O`:

\[ \text{Number of moles} = \frac{\text{mass of solute}}{\text{molar mass}} = \frac{5.6 \, \text{g}}{315 \, \text{g/mol}} \approx 0.01778 \, \text{mol} \]

- The mass of the solvent (water) is 100 g, which is 0.1 kg. Now we can calculate the molality:

\[ \text{Molality} = \frac{0.01778 \, \text{mol}}{0.1 \, \text{kg}} = 0.1778 \, \text{mol/kg} \approx 0.18 \, \text{mol/kg} \]

### Step 3: Calculate the molality of hydroxide ions
- The dissociation of `Ba(OH)2` in solution can be represented as:

\[ Ba(OH)2 \rightarrow Ba^{2+} + 2OH^{-} \]

- From this equation, we see that 1 mole of `Ba(OH)2` produces 2 moles of `OH^-` ions. Therefore, the molality of hydroxide ions will be:

\[ \text{Molality of } OH^{-} = 2 \times \text{Molality of } Ba(OH)2 \approx 2 \times 0.18 \approx 0.36 \, \text{mol/kg} \]

### Final Answer
The molality of hydroxide ions in the saturated solution of `Ba(OH)2 · 8H2O` at `288 K` is approximately **0.36 mol/kg**.

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