What is the `[OH^(-)]` in the final solution prepared by mixing `20.0 mL` of `0.050 M HCl` with `30.0 mL` of `0.10 M Ba(OH)_(2)`?

Options

A. 0.12 M
B. 0.10 M
C. 0.40 M
D. 0.0050 M

Answer

To find the concentration of hydroxide ions \([OH^-]\) in the final solution prepared by mixing \(20.0 \, \text{mL}\) of \(0.050 \, \text{M} \, \text{HCl}\) with \(30.0 \, \text{mL}\) of \(0.10 \, \text{M} \, \text{Ba(OH)}_2\), we can follow these steps:

### Step 1: Calculate the number of moles of HCl
To find the number of moles of HCl, we use the formula:
\[ \text{Number of moles} = \text{Volume (L)} \times \text{Molarity (M)} \]
Convert \(20.0 \, \text{mL}\) to liters:
\[ 20.0 \, \text{mL} = 0.0200 \, \text{L} \]
Now, calculate the moles of HCl:
\[ \text{Moles of HCl} = 0.0200 \, \text{L} \times 0.050 \, \text{M} = 0.00100 \, \text{moles} \, (1 \, \text{mmol}) \]

### Step 2: Calculate the number of moles of Ba(OH)₂
Now, calculate the number of moles of Ba(OH)₂ using the same formula. Convert \(30.0 \, \text{mL}\) to liters:
\[ 30.0 \, \text{mL} = 0.0300 \, \text{L} \]
Now, calculate the moles of Ba(OH)₂:
\[ \text{Moles of Ba(OH)}_2 = 0.0300 \, \text{L} \times 0.10 \, \text{M} = 0.00300 \, \text{moles} \, (3 \, \text{mmol}) \]

### Step 3: Determine the moles of hydroxide ions produced
Each mole of Ba(OH)₂ produces 2 moles of hydroxide ions:
\[ \text{Moles of } OH^- = 2 \times \text{Moles of Ba(OH)}_2 = 2 \times 0.00300 \, \text{moles} = 0.00600 \, \text{moles} \, (6 \, \text{mmol}) \]

### Step 4: Neutralization reaction
The reaction between HCl and Ba(OH)₂ can be represented as:
\[ \text{HCl} + \text{Ba(OH)}_2 \rightarrow \text{BaCl}_2 + 2 \text{H}_2\text{O} \]
From the stoichiometry of the reaction, 1 mole of HCl reacts with 2 moles of \(OH^-\).

Since we have:
- \(1 \, \text{mmol} \, \text{HCl}\) (which provides \(1 \, \text{mmol} \, H^+\))
- \(6 \, \text{mmol} \, OH^-\)

The \(1 \, \text{mmol} \, H^+\) will neutralize \(1 \, \text{mmol} \, OH^-\), leaving:
\[ \text{Remaining } OH^- = 6 \, \text{mmol} - 1 \, \text{mmol} = 5 \, \text{mmol} \]

### Step 5: Calculate the final concentration of hydroxide ions
The total volume of the solution after mixing is:
\[ 20.0 \, \text{mL} + 30.0 \, \text{mL} = 50.0 \, \text{mL} = 0.0500 \, \text{L} \]
Now, calculate the concentration of hydroxide ions:
\[ [OH^-] = \frac{\text{Moles of } OH^-}{\text{Total Volume (L)}} = \frac{0.00500 \, \text{moles}}{0.0500 \, \text{L}} = 0.100 \, \text{M} \]

### Final Answer
The concentration of hydroxide ions \([OH^-]\) in the final solution is \(0.100 \, \text{M}\).

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