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nexus9112 [7]
3 years ago
8

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
Chemistry
2 answers:
Mnenie [13.5K]3 years ago
8 0

Answer:

1, 3, and 4

Step-by-step explanation:

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

<em>1) HCl </em>

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

<em>2) H₂C₂O₄ </em>

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

<em>3) HC₂H₃O₂ </em>

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

<em>4) HBr </em>

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.

serious [3.7K]3 years ago
5 0

Answer:

The answer is: HCl, acetic acid and HBr

Explanation:

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

0.1\frac{mol}{L} *25mL*\frac{1L}{1000mL} =0.0025mol

The moles of HCl is:

0.1\frac{mol}{L} *25mL*\frac{1L}{1000mL} =0.0025mol

The moles of oxalic acid is:

0.1\frac{mol}{L} *50mL*\frac{1L}{1000mL} =0.005moles

The moles of acetic acid is:

0.1\frac{mol}{L} *25mL*\frac{1L}{1000mL} =0.0025mol

The moles of HBr is:

0.2\frac{mol}{L} *12.5mL*\frac{1L}{1000mL} =0.0025moles

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