Question

In which of the titrations described below will the first (or only) equivalence point be reached upon the addition of 25.0 mL of 0.100 M NaOH? 1) The titration of 25.0 mL of 0.100 M HCl (hydrochloric acid) 2) The titration of 50.0 mL of 0.100 M H2C2O4 (oxalic acid) 3) The titration of 25.0 mL of 0.100 M HC2H3O2 (acetic acid) 4) The titration of 12.5 mL of 0.200 M HBr (hydrobromic acid) View Available Hint(s) In which of the titrations described below will the first (or only) equivalence point be reached upon the addition of 25.0 mL of 0.100 M NaOH? 1) The titration of 25.0 mL of 0.100 M HCl (hydrochloric acid) 2) The titration of 50.0 mL of 0.100 M H2C2O4 (oxalic acid) 3) The titration of 25.0 mL of 0.100 M HC2H3O2 (acetic acid) 4) The titration of 12.5 mL of 0.200 M HBr (hydrobromic acid) 1 only 1, 3, and 4 both 1 and 3 All four titrations will have their first (or only) equivalence point after the addition of 25.0 mL of 0.100 M NaOH

Answer 1

The answer is: HCl, acetic acid and HBr

Step-by-step explanation:

In the equivalence point the moles of acid and base are the same, therefore we have:

`0.1(mol)/(L) *25mL*(1L)/(1000mL) =0.0025mol`

The moles of HCl is:

`0.1(mol)/(L) *25mL*(1L)/(1000mL) =0.0025mol`

The moles of oxalic acid is:

`0.1(mol)/(L) *50mL*(1L)/(1000mL) =0.005moles`

The moles of acetic acid is:

`0.1(mol)/(L) *25mL*(1L)/(1000mL) =0.0025mol`

The moles of HBr is:

`0.2(mol)/(L) *12.5mL*(1L)/(1000mL) =0.0025moles`

Answer 2

1, 3, and 4

Step-by-step explanation:

We must calculate the volume of NaOH needed for each titration.

1) HCl

HCl + NaOH ⟶ NaCl + H₂O

n(HCl) = 25.0 mL × (0.100 mmol/1mL) = 2.50 mmol

n(NaOH) = 2.50 mmol HCl × (1 mmol NaOH/1 mmol HCl

= 2.50 mmol NaOH

V(NaOH) = 2.50 mmol × (1 mL/0.100 mmol) = 25.0 mL

2) H₂C₂O₄

H₂C₂O₄ + NaOH ⟶ NaHC₂O₄

n(H₂C₂O₄) = 50.0 mL × (0.100 mmol/1mL) = 5.00 mmol

n(NaOH) = 5.00 mmol H₂C₂O₄ × (1 mmol NaOH/1 mmol H₂C₂O₄)

= 5.00 mmol NaOH

V(NaOH) = 5.00 mmol × (1 mL/0.100 mmol) = 50.0 mL

3) HC₂H₃O₂

HC₂H₃O₂ + NaOH ⟶ NaC₂H₃O₂ + H₂O

n(HC₂H₃O₂) = 25.0 mL × (0.100 mmol/1mL) = 2.50 mmol

n(NaOH) = 2.50 mmol HCl × (1 mmol NaOH/1 mmol HCl)

= 2.50 mmol NaOH

V(NaOH) = 2.50 mmol × (1 mL/0.100 mmol) = 25.0 mL

4) HBr

HBr + NaOH ⟶ NaBr + H₂O

n(HBr) = 25.0 mL × (0.100 mmol/1mL) =2.50 mmol

n(NaOH) = 2.50 mmol HCl × (1 mmol NaOH/1 mmol HCl)

= 2.50 mmol NaOH

V(NaOH) = 2.50 mmol × (1 mL/0.100 mmol) = 25.0 mL

Titrations 1, 3, and 4 reach the first or only equivalence point at 25.0 mL NaOH.