自然科學
大學
看不懂老師的算式 有人能幫我拆解步驟解釋一下嗎
或是另外寫易懂的算式過程
拜偷
A sample containing 25.14g of neutral salts, glucose and a sodium carbonate/bicarbonate
buffer was dissolved in 100 ml of water. A 25 ml aliquot of the resultant solution required
20.35 ml of 0.0987 M HCl when titrated to the PP end-point. A second 25 ml aliquot was
titrated to the MO end-point and required 56.75 ml of the acid. Calculate the percentage of
Na2CO3 (molecular weight 106) and NaHCO3 (molecular weight 84) in the sample.
含有25.14克中性鹽、葡萄糖和碳酸鈉/碳酸氫鈉的樣品將緩衝液溶解在100ml水中。需要25毫升等份所得溶液滴定至PP終點
時,使用20.35ml 0.0987 M HCl。第二份25毫升等分樣本是滴定至MO終點,需要56.75ml酸。計算百分比樣品中的Na2C03
(分子量106)和NaHCO3(分子量84)
39
Naz CO3
PP
Mo
7/2003
✓
→ NaHCO3
(20.35×0.0987) x 106
Na2CO3 =
x100%
25.14 x 25 x 1000
100
NaHCO3 =
(56.75- 2x 30.35) × 84
25.14×100
* 100%
解答
尚無回答
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