Explanation:
To neutralize sulfuric acid (H2SO4) with potassium hydroxide (KOH), we first need to balance the chemical equation for the neutralization reaction:
H2SO4(aq) + 2KOH(aq) → K2SO4(aq) + 2H2O(l)
From the balanced equation, we can see that 1 mole of sulfuric acid reacts with 2 moles of potassium hydroxide.
Given that the concentration of both the sulfuric acid and the potassium hydroxide is 0.200 mol/dm³ and the volume of sulfuric acid is 25.0 cm³ (which is equivalent to 0.025 dm³), we can use the formula:
\[ \text{Volume of solution} = \frac{\text{Number of moles}}{\text{Concentration}} \]
1. **Calculate the number of moles of sulfuric acid:**
\[ \text{Number of moles} = \text{Concentration} \times \text{Volume} \]
\[ \text{Number of moles} = 0.200 \, \text{mol/dm³} \times 0.025 \, \text{dm³} = 0.005 \, \text{mol} \]
2. **Determine the number of moles of potassium hydroxide required:**
Since the molar ratio between sulfuric acid and potassium hydroxide is 1:2, we need twice as many moles of potassium hydroxide as sulfuric acid.
\[ \text{Number of moles of KOH} = 2 \times \text{Number of moles of H}_2\text{SO}_4 = 2 \times 0.005 \, \text{mol} = 0.010 \, \text{mol} \]
3. **Calculate the volume of potassium hydroxide solution required:**
\[ \text{Volume of solution} = \frac{\text{Number of moles}}{\text{Concentration}} \]
\[ \text{Volume of solution} = \frac{0.010 \, \text{mol}}{0.200 \, \text{mol/dm}³} = 0.050 \, \text{dm}³ \]
So, 0.050 dm³ or 50.0 cm³ of 0.200 mol/dm³ potassium hydroxide solution is required to neutralize 25.0 cm³ of 0.200 mol/dm³ aqueous sulfuric acid.
