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.
Need a fast expert's response?Submit order
and get a quick answer at the best price
for any assignment or question with DETAILED EXPLANATIONS!
You're welcome. We are glad to be helpful.
If you really liked our service please press like-button beside answer field. Thank you!
Thank you for answering my question!!!