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.
