25 mL of acetic acid solution of an unknown concentration was titrated to the equivalence point with 11.25 mL of 0.998 M NaOH.?
2017-04-12 1:47 am
25 mL of acetic acid solution of an unknown concentration was titrated to the equivalence point with 11.25 mL of 0.998 M NaOH. What is the molarity of the acid solution?
回答 (1)
2017-04-12 1:56 am
Equation for the reaction :
CH₃COOH + NaOH → CH₃COONa + H₂O
Mole ratio CH₃COOH : NaOH = 1 : 1
No. of moles of NaOH reacted = (0.998 mol/L) × (11.25/1000 L) = 0.01123 mol
No. of moles of CH₃COOH reacted = (0.01123 mol) × 1 = 0.01123 mol
Molarity of CH₃COOH = (0.01123 mol) / (25/1000) = 0.449 M
CH₃COOH + NaOH → CH₃COONa + H₂O
Mole ratio CH₃COOH : NaOH = 1 : 1
No. of moles of NaOH reacted = (0.998 mol/L) × (11.25/1000 L) = 0.01123 mol
No. of moles of CH₃COOH reacted = (0.01123 mol) × 1 = 0.01123 mol
Molarity of CH₃COOH = (0.01123 mol) / (25/1000) = 0.449 M
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