# Answer to Question #21147 in Organic Chemistry for Sadaam Ahmed

Question #21147

What volume of 2.0 mol/L sulfuric acid will be exactly neutralized by 12.3 mL of 1.5 mol/L potassium hydroxide?

Expert's answer

c = n / V

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

H

According to reaction above one mol of sulfuric acid reacts with two mol of potassium hydroxide. Thus, the amount of sulfuric acid is

n(H

The volume of sulfuric acid solution is

V(H

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.

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

H

_{2}SO_{4}+ 2KOH = K_{2}SO_{4}+ 2H_{2}OAccording to reaction above one mol of sulfuric acid reacts with two mol of potassium hydroxide. Thus, the amount of sulfuric acid is

n(H

_{2}SO_{4}) = n(KOH) / 2 = 18.45 / 2 = 9.225 mmolThe volume of sulfuric acid solution is

V(H

_{2}SO_{4}) = n(H_{2}SO_{4}) / c(H_{2}SO_{4}) = 9.225 * 10^-3 mol / 2.0 mol/L = 0.0046 L = 4.6 mLAnswer: 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.

## Comments

Assignment Expert08.01.13, 15:09You're welcome. We are glad to be helpful.

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Sadaam Ahmed29.12.12, 22:13Thank you for answering my question!!!

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