Aspirin (C9H8O4) is an acid which can be titrated with a base to determine purity.?
2020-08-17 3:29 pm
If an aspirin tablet weighing 1.83 g is titrated with standardized 0.2158 M KOH, the endpoint is reached after 24.42 mL of KOH has been added. What is the percent aspirin in the tablet?
回答 (1)
2020-08-17 10:08 pm
Molar mass of aspirin (C₉H₈O₄) = (12.0×9 + 1.0×8 + 16.0×4) g/mol = 180.0 g/mol
The equation for the reaction between aspirin and KOH is shown in the figure below.
The equation can be simplified as:
aspirin + 2KOH → products
Mole ratio aspirin : KOH = 1 : 2
Moles of KOH reacted = (0.2158 mol/L) × (24.42/1000 L) = 0.005270 mol
Moles of aspirin reacted = (0.005270 mol) × (1/2) = 0.002635 mol
Mass of aspirin reacted = (0.002635 mol) × (180 g/mol) = 0.474 g
% aspirin in the tablet = (0.474/1.83) × 100% = 25.9%
====
OR:
(0.2158 mol KOH / 1000 mL KOH solution) × (24.42 mL KOH solution) × (1 mol aspirin / 2 mol KOH) × (180.0 g aspirin / 1 mol aspirin)
= 0.474 g aspirin
(0.474 g aspirin / 1.83 g tablet) × 100%
= 25.9% aspirin/tablet
The equation for the reaction between aspirin and KOH is shown in the figure below.
The equation can be simplified as:
aspirin + 2KOH → products
Mole ratio aspirin : KOH = 1 : 2
Moles of KOH reacted = (0.2158 mol/L) × (24.42/1000 L) = 0.005270 mol
Moles of aspirin reacted = (0.005270 mol) × (1/2) = 0.002635 mol
Mass of aspirin reacted = (0.002635 mol) × (180 g/mol) = 0.474 g
% aspirin in the tablet = (0.474/1.83) × 100% = 25.9%
====
OR:
(0.2158 mol KOH / 1000 mL KOH solution) × (24.42 mL KOH solution) × (1 mol aspirin / 2 mol KOH) × (180.0 g aspirin / 1 mol aspirin)
= 0.474 g aspirin
(0.474 g aspirin / 1.83 g tablet) × 100%
= 25.9% aspirin/tablet
收錄日期: 2021-04-24 07:57:37
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20200817072916AATc5vh