What volume of 0.10 M `H_(2)SO_(4)` must be added to 50 mL of a 0.10 NaOH solution to make a solution in which molarity of the `H_(2)SO_(4)` is 0.050M?

Options

A. 400mL
B. 200mL
C. 100mL
D. none of these

Answer

To solve the problem of determining what volume of 0.10 M \( H_2SO_4 \) must be added to 50 mL of a 0.10 M NaOH solution to achieve a final molarity of 0.050 M \( H_2SO_4 \), we can follow these steps:

### Step 1: Understand the Reaction
The balanced chemical reaction between sulfuric acid (\( H_2SO_4 \)) and sodium hydroxide (\( NaOH \)) is:
\[ H_2SO_4 + 2 NaOH \rightarrow Na_2SO_4 + 2 H_2O \]
From this reaction, we see that 1 mole of \( H_2SO_4 \) reacts with 2 moles of \( NaOH \).

### Step 2: Calculate Moles of NaOH
We have a 0.10 M NaOH solution and a volume of 50 mL. First, convert the volume from mL to L:
\[ 50 \, \text{mL} = 0.050 \, \text{L} \]
Now, calculate the moles of \( NaOH \):
\[ \text{Moles of } NaOH = \text{Molarity} \times \text{Volume} = 0.10 \, \text{mol/L} \times 0.050 \, \text{L} = 0.005 \, \text{mol} \]

### Step 3: Determine Moles of \( H_2SO_4 \) Needed
From the reaction stoichiometry, we know that 1 mole of \( H_2SO_4 \) reacts with 2 moles of \( NaOH \). Therefore, the moles of \( H_2SO_4 \) needed to react with 0.005 moles of \( NaOH \) is:
\[ \text{Moles of } H_2SO_4 = \frac{0.005 \, \text{mol NaOH}}{2} = 0.0025 \, \text{mol} \]

### Step 4: Calculate the Volume of \( H_2SO_4 \) Required
We want the final molarity of \( H_2SO_4 \) to be 0.050 M. Using the formula for molarity:
\[ \text{Molarity} = \frac{\text{Moles}}{\text{Volume (L)}} \]
We can rearrange this to find the volume:
\[ \text{Volume (L)} = \frac{\text{Moles}}{\text{Molarity}} = \frac{0.0025 \, \text{mol}}{0.050 \, \text{mol/L}} = 0.050 \, \text{L} \]
Convert this volume to mL:
\[ 0.050 \, \text{L} = 50 \, \text{mL} \]

### Conclusion
The volume of 0.10 M \( H_2SO_4 \) that must be added is **50 mL**.

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