50 mL of `H_(2)O` is added to `50 mL` of `1 xx 10^(-3)M` barium hydroxide solution. What is the pH of the resulting solution?

Options

A. `3.0`
B. `3.3`
C. `11.0`
D. `11.7`

Explanation

To find the pH of the resulting solution when 50 mL of water is added to 50 mL of a 1 x 10^(-3) M barium hydroxide solution, we can follow these steps:

### Step 1: Determine the initial conditions
We have:
- Volume of barium hydroxide solution (V1) = 50 mL
- Molarity of barium hydroxide solution (M1) = 1 x 10^(-3) M

### Step 2: Calculate the total volume of the resulting solution
When 50 mL of water is added to the barium hydroxide solution, the total volume (V2) becomes:
\[ V2 = V1 + \text{Volume of water} = 50 \, \text{mL} + 50 \, \text{mL} = 100 \, \text{mL} \]

### Step 3: Use the dilution formula to find the new molarity
The dilution formula is given by:
\[ M1 \times V1 = M2 \times V2 \]
Where:
- M2 = new molarity of the barium hydroxide solution after dilution

Substituting the known values:
\[ (1 \times 10^{-3} \, \text{M}) \times (50 \, \text{mL}) = M2 \times (100 \, \text{mL}) \]

### Step 4: Solve for M2
Rearranging the equation to find M2:
\[ M2 = \frac{(1 \times 10^{-3} \, \text{M}) \times (50 \, \text{mL})}{100 \, \text{mL}} \]
\[ M2 = \frac{50 \times 10^{-3}}{100} = 0.5 \times 10^{-3} \, \text{M} \]

### Step 5: Determine the concentration of hydroxide ions
Barium hydroxide (Ba(OH)₂) dissociates in water as follows:
\[ \text{Ba(OH)}_2 \rightarrow \text{Ba}^{2+} + 2 \text{OH}^- \]
This means that for every mole of barium hydroxide, we get 2 moles of hydroxide ions. Therefore, the concentration of hydroxide ions \([OH^-]\) is:
\[ [OH^-] = 2 \times M2 = 2 \times (0.5 \times 10^{-3} \, \text{M}) = 1 \times 10^{-3} \, \text{M} \]

### Step 6: Calculate the pOH
The pOH can be calculated using the formula:
\[ pOH = -\log[OH^-] \]
Substituting the value of \([OH^-]\):
\[ pOH = -\log(1 \times 10^{-3}) \]
Using the logarithmic property:
\[ pOH = 3 \]

### Step 7: Calculate the pH
We know that:
\[ pH + pOH = 14 \]
Thus, we can find the pH:
\[ pH = 14 - pOH = 14 - 3 = 11 \]

### Final Answer
The pH of the resulting solution is **11**.

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