Neutralization by Acid-Base Titration Problems HELP

[calculus_love]

Neutralization by Acid-Base Titration Problems !HELP!

Hey guys, could you please take a look at my work and let me know if I did it correctly or not? I'm not sure if I did, so if I did anything wrong I'd really appreciate if you could explain! Thanks!

1. A sample of solid potassium hydroxide, KOH, weighing 15.0 grams is dissolved in 75.0 mL of deionized water to make a solution. What volume in mL of 0.235 M sulfuric acid solution will neutralize this solution? (Hint: write a balanced chemical equation first.)
2KOH(aq) + H2SO4(aq) = K2SO4(aq) + 2H2O(l)

15.0g KOH × (1 mol KOH / 56.11g KOH) × (1 mol H2SO4 / 2 mol KOH) × (1 L H2SO4(aq)/0.235 mol H2SO4) × (1 mL / 10^-3 L) = 568 L

2. What volume in milliliters of 0.500 M HNO3 is required to neutralize 40.00 milliliters of a 0.200 M NaOH solution?
NaOH(aq) + HNO3(aq) = NaNO3(aq) + H2O(l)

40.00mL NaOH(aq) × (10^-3 L / 1 mL) × (0.200 mol NaOH / 1 L NaOH(aq)) × (1 mol HNO3 / 1 mol NaOH) × (1 L HNO3(aq) / 0.500 mol HNO3) × (1 mL / 10^-3 L) = 240 mL

3. How many grams of calcium hydroxide, Ca(OH)2 are required to neutralize 47.76 milliliters of 0.750 M H2SO4 solution?
Ca(OH)2(aq) + H2SO4(aq) = CaSO4(aq) + 2H2O(l)

47.76 mL H2SO4(aq) × (10^-3 L / 1 mL) × (0.750 mol H2SO4 / 1 L H2SO4(aq)) × (1 mol Ca(OH)2 / 1 mol H2SO4) × (74.10g Ca(OH)2 / 1 mol Ca(OH)2) = 2.65 g

 

[chemisttree]

Hey guys, could you please take a look at my work and let me know if I did it correctly or not? I'm not sure if I did, so if I did anything wrong I'd really appreciate if you could explain! Thanks!

1. A sample of solid potassium hydroxide, KOH, weighing 15.0 grams is dissolved in 75.0 mL of deionized water to make a solution. What volume in mL of 0.235 M sulfuric acid solution will neutralize this solution? (Hint: write a balanced chemical equation first.)
2KOH(aq) + H2SO4(aq) = K2SO4(aq) + 2H2O(l)

15.0g KOH × (1 mol KOH / 56.11g KOH) × (1 mol H2SO4 / 2 mol KOH) × (1 L H2SO4(aq)/0.235 mol H2SO4) × (1 mL / 10^-3 L) = 568 L

Units are wrong.

2. What volume in milliliters of 0.500 M HNO3 is required to neutralize 40.00 milliliters of a 0.200 M NaOH solution?
NaOH(aq) + HNO3(aq) = NaNO3(aq) + H2O(l)

40.00mL NaOH(aq) × (10^-3 L / 1 mL) × (0.200 mol NaOH / 1 L NaOH(aq)) × (1 mol HNO3 / 1 mol NaOH) × (1 L HNO3(aq) / 0.500 mol HNO3) × (1 mL / 10^-3 L) = 240 mL

Obviously wrong. Your answer indicates that you will need six times the volume of the weaker solution to neutralize using the stronger solution. By inspection you should expect to use less volume of the stronger solution to neutralize a given volume of the weaker solution.