A 20.00 ml sample of `Ba (OH)_2` solution is titrated with 0.245 M HCI. If 27.15 ml of HCI is required, then the molarity of the `Ba (OH)_2` solution will be :

Options

A. 0.166 M
B. 0.180 M
C. 0.333 M
D. 0.666 M

Answer

To find the molarity of the barium hydroxide (Ba(OH)₂) solution, we can follow these steps:

### Step 1: Write the balanced chemical equation
When barium hydroxide reacts with hydrochloric acid (HCl), the reaction can be represented as:
\[ \text{Ba(OH)}_2 + 2 \text{HCl} \rightarrow \text{BaCl}_2 + 2 \text{H}_2\text{O} \]

### Step 2: Determine the mole ratio
From the balanced equation, we see that 1 mole of Ba(OH)₂ reacts with 2 moles of HCl. Therefore, the mole ratio is:
\[ \text{Ba(OH)}_2 : \text{HCl} = 1 : 2 \]

### Step 3: Calculate the moles of HCl used
We know the molarity (M) of HCl and the volume (V) used in the titration:
- Molarity of HCl = 0.245 M
- Volume of HCl = 27.15 mL = 0.02715 L (since 1 mL = 0.001 L)

Using the formula:
\[ \text{Moles of HCl} = \text{Molarity} \times \text{Volume} \]
\[ \text{Moles of HCl} = 0.245 \, \text{mol/L} \times 0.02715 \, \text{L} \]
\[ \text{Moles of HCl} = 0.00665 \, \text{moles} \]

### Step 4: Calculate the moles of Ba(OH)₂
From the mole ratio, we know that 1 mole of Ba(OH)₂ reacts with 2 moles of HCl. Therefore, the moles of Ba(OH)₂ can be calculated as:
\[ \text{Moles of Ba(OH)}_2 = \frac{1}{2} \times \text{Moles of HCl} \]
\[ \text{Moles of Ba(OH)}_2 = \frac{1}{2} \times 0.00665 \]
\[ \text{Moles of Ba(OH)}_2 = 0.003325 \, \text{moles} \]

### Step 5: Calculate the molarity of Ba(OH)₂
The molarity (M) of Ba(OH)₂ can be calculated using the formula:
\[ \text{Molarity} = \frac{\text{Moles}}{\text{Volume in L}} \]
The volume of Ba(OH)₂ solution used is 20.00 mL = 0.02000 L.

Thus,
\[ \text{Molarity of Ba(OH)}_2 = \frac{0.003325 \, \text{moles}}{0.02000 \, \text{L}} \]
\[ \text{Molarity of Ba(OH)}_2 = 0.16625 \, \text{M} \]

### Final Answer
The molarity of the barium hydroxide solution is approximately:
\[ \text{Molarity of Ba(OH)}_2 \approx 0.1663 \, \text{M} \]

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