# Sample problems acid-base titration

1. Problem: How many milliliters of 0.500 M NaOH are needed to neutralize 20.0 mL of 0.300 M HCl? Solution: Moles of HCl = 0.300 M x 20.0 mL / 1000 = 0.006 mol; Volume of NaOH = 0.006 mol / 0.500 M = 0.012 L = 12.0 mL

2. Problem: How many milliliters of 0.200 M KOH are needed to neutralize 30.0 mL of 0.250 M HNO₃? Solution: Moles of HNO₃ = 0.250 M x 30.0 mL / 1000 = 0.0075 mol; Volume of KOH = 0.0075 mol / 0.200 M = 0.0375 L = 37.5 mL
3. Problem: How many milliliters of 0.350 M H₂SO₄ are needed to neutralize 25.0 mL of 0.400 M NaOH? Solution: Moles of NaOH = 0.400 M x 25.0 mL / 1000 = 0.010 mol; Volume of H₂SO₄ = 0.010 mol / (2 x 0.350 M) = 0.01429 L = 14.29 mL
4. Problem: How many milliliters of 0.600 M HCl are needed to neutralize 50.0 mL of 0.500 M KOH? Solution: Moles of KOH = 0.500 M x 50.0 mL / 1000 = 0.025 mol; Volume of HCl = 0.025 mol / 0.600 M = 0.04167 L = 41.67 mL
5. Problem: How many milliliters of 0.750 M NaOH are needed to neutralize 40.0 mL of 0.600 M HNO₃? Solution: Moles of HNO₃ = 0.600 M x 40.0 mL / 1000 = 0.024 mol; Volume of NaOH = 0.024 mol / 0.750 M = 0.032 L = 32.0 mL
6. Problem: How many milliliters of 0.500 M H₂SO₄ are needed to neutralize 35.0 mL of 0.450 M KOH? Solution: Moles of KOH = 0.450 M x 35.0 mL / 1000 = 0.01575 mol; Volume of H₂SO₄ = 0.01575 mol / (2 x 0.500 M) = 0.01575 L = 15.75 mL
7. Problem: How many milliliters of 0.400 M NaOH are needed to neutralize 45.0 mL of 0.350 M HCl? Solution: Moles of HCl = 0.350 M x 45.0 mL / 1000 = 0.01575 mol; Volume of NaOH = 0.01575 mol / 0.400 M = 0.039375 L = 39.375 mL
8. Problem: How many milliliters of 0.300 M KOH are needed to neutralize 60.0 mL of 0.250 M HNO₃? Solution: Moles of HNO₃ = 0.250 M x 60.0 mL / 1000 = 0.015 mol; Volume of KOH = 0.015 mol / 0.300 M = 0.05 L = 50.0 mL
9. Problem: A titration is performed on 25.0 mL of 0.150 M HCl with 0.100 M NaOH. How many mL of NaOH are needed to reach the equivalence point?

Solution: Since the reaction between NaOH and HCl goes in a 1:1 ratio, we can use the formula n₁V₁ = n₂V₂, where n is the normality and V is the volume. Then we find that V₂ = (n₁V₁) / n₂ = (0.150 mol/L x 25.0 mL) / 0.100 mol/L = 37.5 mL.

10. Problem: How many mL of 0.200 M H₂SO₄ are needed to neutralize 50.0 mL of 0.300 M KOH?

Solution: H₂SO₄ reacts with KOH in a 1:2 ratio. So, V₂ = 2n₁V₁ / n₂ = 2 x 0.300 mol/L x 50.0 mL / 0.200 mol/L = 150 mL.

11. Problem: What is the molarity of a HNO₃ solution if 40.0 mL of the acid neutralizes 35.0 mL of 0.400 M NaOH?

Solution: The reaction ratio between HNO₃ and NaOH is 1:1. So, n₁ = n₂V₂ / V₁ = (0.400 mol/L x 35.0 mL) / 40.0 mL = 0.350 M.

12. Problem: How many grams of NaOH (Molar mass = 40.00 g/mol) are needed to neutralize 20.0 mL of 1.50 M HCl?

Solution: Moles of HCl = Molarity x Volume = 1.50 mol/L x 0.020 L = 0.030 mol. Since NaOH and HCl react in a 1:1 ratio, moles of NaOH = 0.030 mol. So, grams of NaOH = moles x molar mass = 0.030 mol x 40.00 g/mol = 1.20 g.

13. Problem: Calculate the molarity of HCl if it took 25.0 mL of 0.250 M NaOH to neutralize 10.0 mL of HCl.

Solution: The reaction ratio between HCl and NaOH is 1:1. So, n₁ = n₂V₂ / V₁ = (0.250 mol/L x 25.0 mL) / 10.0 mL = 0.625 M.

14. Problem: A 50.0 mL solution of 0.200 M H₂SO₄ is titrated with a 0.400 M KOH solution. What volume of KOH is needed to reach the equivalence point?

Solution: The reaction ratio between H₂SO₄ and KOH is 1:2. So, V₂ = n₁V₁ / (n₂/2) = 0.200 mol/L x 50.0 mL / (0.400 mol/L / 2) = 50.0 mL.

15. Problem: What volume of 0.100 M HCl is required to neutralize 25.0 mL of 0.200 M Ba(OH)₂?

Solution: The reaction ratio between HCl and Ba(OH)₂ is 2:1. So, V₁ = 2n₂V₂ / n₁ = 2 x 0.200 mol/L x 25.0 mL / 0.100 mol/L = 100 mL.

16. Problem: What is the molarity of a NaOH solution if 20.0 mL of the base neutralize 15.0 mL of 0.500 M HCl?

Solution: The reaction ratio between NaOH and HCl is 1:1. So, n₂ = n₁V₁ / V₂ = 0.500 mol/L x 15.0 mL / 20.0 mL = 0.375 M.

17. Problem: Calculate the molarity of HNO₃ if it took 30.0 mL of 0.300 M KOH to neutralize 20.0 mL of HNO₃.

Solution: The reaction ratio between HNO₃ and KOH is 1:1. So, n₁ = n₂V₂ / V₁ = 0.300 mol/L x 30.0 mL / 20.0 mL = 0.450 M.

18. Problem: A 45.0 mL solution of 0.150 M H₂SO₄ is titrated with a 0.300 M NaOH solution. What volume of NaOH is needed to reach the equivalence point?

Solution: The reaction ratio between H₂SO₄ and NaOH is 1:2. So, V₂ = n₁V₁ / (n₂/2) = 0.150 mol/L x 45.0 mL / (0.300 mol/L / 2) = 45.0 mL.

19. Problem: How many mL of 0.100 M HCl are needed to neutralize 50.0 mL of 0.200 M Ca(OH)₂?

Solution: The reaction ratio between HCl and Ca(OH)₂ is 2:1. So, V₁ = 2n₂V₂ / n₁ = 2 x 0.200 mol/L x 50.0 mL / 0.100 mol/L = 200 mL.