A Level Chemistry - Titration & Half Equivalence Calculations

In a titration, 25.0 cm³ of 0.10 mol dm⁻³ NaOH is required to neutralize 20.0 cm³ of HCl. What is the concentration of HCl?

• 0.080 mol dm⁻³

• 0.125 mol dm⁻³

• 0.075 mol dm⁻³

• 0.150 mol dm⁻³

What volume of 0.50 mol dm⁻³ H₂SO₄ is required to neutralize 50.0 cm³ of 1.00 mol dm⁻³ NaOH?

• 50.0 cm³

• 25.0 cm³

• 100.0 cm³

• 75.0 cm³

50.0 cm³ of 0.10 mol dm⁻³ CH₃COOH is titrated with 0.10 mol dm⁻³ NaOH. What volume of NaOH is required to reach the equivalence point?

• 50.0 cm³

• 100.0 cm³

• 25.0 cm³

• 10.0 cm³

What is the pH at the half-equivalence point of a titration of acetic acid with NaOH, given that the K_a of acetic acid is 1.74 x 10^-5?

• 4.76

• 7.00

• 5.76

• 3.76

25.0 cm³ of 0.20 mol dm⁻³ KOH is used to titrate 20.0 cm³ of HNO₃. What is the concentration of HNO₃?

• 0.250 mol dm⁻³

• 0.200 mol dm⁻³

• 0.150 mol dm⁻³

• 0.125 mol dm⁻³

In a titration of 0.10 mol dm⁻³ HCl with 0.10 mol dm⁻³ NaOH, what is the pH at the equivalence point?

• 3.50

• 7.00

• 9.00

• 12.00

How many moles of NaOH are needed to completely neutralize 25.0 cm³ of 0.20 mol dm⁻³ H₂SO₄?

• 0.0100 mol

• 0.0050 mol

• 0.0200 mol

• 0.0150 mol

During a titration, 40.0 cm³ of 0.10 mol dm⁻³ NaOH neutralizes 50.0 cm³ of H₃PO₄. What is the concentration of H₃PO₄?

• 0.027 mol dm⁻³

• 0.080 mol dm⁻³

• 0.040 mol dm⁻³

• 0.060 mol dm⁻³

100.0 cm³ of 0.05 mol dm⁻³ NaOH is required to neutralize 50.0 cm³ of H₂SO₄. What is the concentration of H₂SO₄?

• 0.10 mol dm⁻³

• 0.050 mol dm⁻³

• 0.025 mol dm⁻³

• 0.075 mol dm⁻³

50.0 cm³ of 0.100 mol dm⁻³ NaOH is required to neutralize 25.0 cm³ of H₂SO₄. What is the concentration of H₂SO₄?

• 0.100 mol dm⁻³

• 0.050 mol dm⁻³

• 0.200 mol dm⁻³

• 0.150 mol dm⁻³