If a 25.0 mL sample of sulfuric acid is titrated with 50.0 mL of 0.025 M sodium hydroxide to a phenolphthalein endpoint, what is the molarity of the acid?

Options

A. `0.020 M`
B. `0.100 M`
C. `0.025 M`
D. `0.050 M`

Answer

To solve the problem of determining the molarity of sulfuric acid (H₂SO₄) when titrated with sodium hydroxide (NaOH), we can follow these steps:

### Step 1: Write down the information given in the problem.
- Volume of sulfuric acid (V₁) = 25.0 mL
- Volume of sodium hydroxide (V₂) = 50.0 mL
- Molarity of sodium hydroxide (M₂) = 0.025 M

### Step 2: Convert volumes from mL to L.
- V₁ = 25.0 mL = 0.025 L
- V₂ = 50.0 mL = 0.050 L

### Step 3: Write the balanced chemical equation for the reaction.
The balanced equation for the reaction between sulfuric acid and sodium hydroxide is:
\[ \text{H}_2\text{SO}_4 + 2 \text{NaOH} \rightarrow \text{Na}_2\text{SO}_4 + 2 \text{H}_2\text{O} \]

From the equation, we see that 1 mole of sulfuric acid reacts with 2 moles of sodium hydroxide.

### Step 4: Use the titration formula.
Using the formula for titration:
\[ M₁ \times V₁ = M₂ \times V₂ \]
Where:
- M₁ = molarity of sulfuric acid (what we want to find)
- V₁ = volume of sulfuric acid in L
- M₂ = molarity of sodium hydroxide
- V₂ = volume of sodium hydroxide in L

### Step 5: Substitute the known values into the equation.
Since sulfuric acid provides 2 moles of H⁺ ions for every mole of H₂SO₄, we need to account for this in our calculations. Therefore, we can express the relationship as:
\[ M₁ \times V₁ = \frac{M₂ \times V₂}{2} \]

Substituting the values:
\[ M₁ \times 0.025 = \frac{0.025 \times 0.050}{2} \]

### Step 6: Calculate the right side of the equation.
Calculating the right side:
\[ 0.025 \times 0.050 = 0.00125 \]
Now divide by 2:
\[ \frac{0.00125}{2} = 0.000625 \]

### Step 7: Solve for M₁.
Now we can solve for M₁:
\[ M₁ \times 0.025 = 0.000625 \]
\[ M₁ = \frac{0.000625}{0.025} \]
\[ M₁ = 0.025 \, \text{M} \]

### Conclusion:
The molarity of the sulfuric acid is **0.025 M**.

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