25.00 mL of a solution containing 10.0 g/L of an unknown acid is titrated?
2017-08-01 11:44 am
25.00 mL of a solution containing 10.0 g/L of an unknown acid is titrated with 0.108 M NaOH. It takes 27.22 mL of NaOH to reach the end point. What is the equivalent weight of the acid? Hint: see the procedure for the lab practical.
回答 (1)
2017-08-01 4:17 pm
1 mole of NaOH dissociates to give 1 mole of OH⁻ ions.
0.108 M NaOH = 0.108 N NaOH = 0.108 eq/L NaOH
No. of equivalents of the unknown acid reacted = No. of equivalents of NaOH reacted
(10 g/L) × (25.00/1000 L) / (Equivalent weight of the acid) = (0.108 eq/L) × (27.22/1000 mL)
Equivalents weight of the acid = 10 × 25.00 / (0.108 × 27.22) g/eq = 85.0 g/eq
0.108 M NaOH = 0.108 N NaOH = 0.108 eq/L NaOH
No. of equivalents of the unknown acid reacted = No. of equivalents of NaOH reacted
(10 g/L) × (25.00/1000 L) / (Equivalent weight of the acid) = (0.108 eq/L) × (27.22/1000 mL)
Equivalents weight of the acid = 10 × 25.00 / (0.108 × 27.22) g/eq = 85.0 g/eq
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