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_{2}SO_{4} + 2KOH = K_{2}SO_{4} + 2H_{2}O

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_{2}SO_{4}) = n(KOH) / 2 = 18.45 / 2 = 9.225 mmol

The 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 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.

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.

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

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