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Suppose [tex]0.09886 \, M \, KOH[/tex] is titrated into [tex]15.00 \, mL \, H_2SO_4[/tex] of unknown concentration until the equivalence point is reached. It takes a total volume of [tex]29.25 \, mL \, KOH[/tex] to neutralize the [tex]H_2SO_4[/tex].

What is the stoichiometric ratio of base to acid?

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To find the stoichiometric ratio of the base (KOH) to the acid (H[tex]\(_2\)[/tex]SO[tex]\(_4\)[/tex]), we need to follow a systematic approach involving the neutralization reaction between KOH and H[tex]\(_2\)[/tex]SO[tex]\(_4\)[/tex]. Let's go through the steps in detail.

### Step 1: Write the balanced chemical equation.
The balanced chemical equation for the reaction between potassium hydroxide (KOH) and sulfuric acid (H[tex]\(_2\)[/tex]SO[tex]\(_4\)[/tex]) is:
[tex]\[ 2 \text{KOH} + \text{H}_2\text{SO}_4 \rightarrow \text{K}_2\text{SO}_4 + 2 \text{H}_2\text{O} \][/tex]

From the equation, 2 moles of KOH react with 1 mole of H[tex]\(_2\)[/tex]SO[tex]\(_4\)[/tex].

### Step 2: Convert the volumes into liters.
- Volume of H[tex]\(_2\)[/tex]SO[tex]\(_4\)[/tex] = 15.00 mL = 0.015 L
- Volume of KOH = 29.25 mL = 0.02925 L

### Step 3: Calculate the moles of KOH.
Using the molarity and the volume of KOH, we can calculate the moles of KOH:
[tex]\[ \text{Moles of KOH} = \text{Molarity} \times \text{Volume} \][/tex]
[tex]\[ \text{Moles of KOH} = 0.09886 \, \text{M} \times 0.02925 \, \text{L} \][/tex]
[tex]\[ \text{Moles of KOH} = 0.002891655 \, \text{moles} \][/tex]

### Step 4: Use the stoichiometric relationship to find moles of H[tex]\(_2\)[/tex]SO[tex]\(_4\)[/tex].
From the balanced equation, we see that 2 moles of KOH react with 1 mole of H[tex]\(_2\)[/tex]SO[tex]\(_4\)[/tex]. Thus, we can calculate the moles of H[tex]\(_2\)[/tex]SO[tex]\(_4\)[/tex] that reacted:
[tex]\[ \text{Moles of H}_2\text{SO}_4 = \frac{\text{Moles of KOH}}{2} \][/tex]
[tex]\[ \text{Moles of H}_2\text{SO}_4 = \frac{0.002891655}{2} \][/tex]
[tex]\[ \text{Moles of H}_2\text{SO}_4 = 0.0014458275 \, \text{moles} \][/tex]

### Step 5: Calculate the stoichiometric ratio.
The stoichiometric ratio of base (KOH) to acid (H[tex]\(_2\)[/tex]SO[tex]\(_4\)[/tex]) is given by:
[tex]\[ \text{Stoichiometric Ratio} = \frac{\text{Moles of KOH}}{\text{Moles of H}_2\text{SO}_4} \][/tex]
[tex]\[ \text{Stoichiometric Ratio} = \frac{0.002891655}{0.0014458275} \][/tex]
[tex]\[ \text{Stoichiometric Ratio} = 2.0 \][/tex]

### Conclusion
The stoichiometric ratio of base (KOH) to acid (H[tex]\(_2\)[/tex]SO[tex]\(_4\)[/tex]) in this neutralization reaction is 2.0.
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