The reaction of sulfuric acid (H2SO4) with potassium hydroxide (KOH) is described by the equation below. Suppose
1 answer:
Answer:
Explanation:
The change of color indicates that the solution was neutralized.
<u>1. Neutralization equation(given):</u>
![H_2SO_4+2KOH\rightarrow K_2SO_4+2H_2O](https://tex.z-dn.net/?f=H_2SO_4%2B2KOH%5Crightarrow%20K_2SO_4%2B2H_2O)
<u>2. Mole ratio:</u>
- 2moles of KOH are neutralized by 1 mol of H₂SO₄
<u>3. Number of moles of H₂SO₄</u>
<u />
- Molarity, M = 0.20M
- Volume, V = 0.017 liter
- M = n/V ⇒ n = M×V =0.20M × 0.017 liter = 0.034mol
<u>4. Number of moles of KOH</u>
![0.034molH_2SO_4\times \dfrac{2molKOH}{1molH_2SO_4}=0.068molKOH](https://tex.z-dn.net/?f=0.034molH_2SO_4%5Ctimes%20%5Cdfrac%7B2molKOH%7D%7B1molH_2SO_4%7D%3D0.068molKOH)
<u>5. Concnetration KOH</u>
- M = n/V = 0.068 mol / 0.06 liter = 1.13 M ← answer
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