Answer to Question #21147 in Organic Chemistry for Sadaam Ahmed
The amount of potassium hydroxide in solution is
n(KOH) = c(KOH) * V(KOH solution) = 1.5 mol/L * 12.3 * 10^-3L = 18.45 * 10^-3 mol = 18.45 mmol
H2SO4 + 2KOH = K2SO4 + 2H2O
According to reaction above one mol of sulfuric acid reacts with two mol of potassium hydroxide. Thus, the amount of sulfuric acid is
n(H2SO4) = n(KOH) / 2 = 18.45 / 2 = 9.225 mmol
The volume of sulfuric acid solution is
V(H2SO4) = n(H2SO4) / c(H2SO4) = 9.225 * 10^-3 mol / 2.0 mol/L = 0.0046 L = 4.6 mL
Answer: 4.6 mL of 2.0 mol/L sulfuric acid will be exactly neutralized by 12.3 mL of 1.5 mol/L potassium hydroxide.
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