Vanuation Question
1 answer:
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The molarity of the acid is 0.086 00 mol/L.
HC₂H₃O₂ + NaOH → NaC₂H₃O₂ + H₂O
Since 1 mol of acid reacts with 1 mol of base, we can use the formula
M₁V₁ = M₂V₂
M₁ = ?; V₁ = 50.00 mL
M₂ = 0.1000 mol/L; V₂ = 43.00 mL
M₁ = M₂ ×
= 0.1000 mol/L × 
= 0.086 00 mol/L
HC₂H₃O₂ + NaOH → NaC₂H₃O₂ + H₂O
Since 1 mol of acid reacts with 1 mol of base, we can use the formula
M₁V₁ = M₂V₂
M₁ = ?; V₁ = 50.00 mL
M₂ = 0.1000 mol/L; V₂ = 43.00 mL
M₁ = M₂ ×