標題:

Titration + dilution calculation~HELP

發問:

A 25 ml aliquot of vinegar was diluted to 250mL. Titration of 50 mL aliquots of the diluted solution required an average of 34.88 ml of 0.096M NaOH. (a) Express the acidity of the vinegar as percent (w/v) acetic acid. Thanks~~~

最佳解答:

Consider the titration between 50 mL diluted vinegar and NaOH. CH3COOH + NaOH → CH3COONa + H2O Mole ratio CH3COOH : NaOH = 1 : 1 No. of moles of NaOH = MV = 0.096 x (34.88/1000) = 0.003348 mol No. of moles of CH3COOH in 50 mL of diluted vinegar = 0.003348 mol No. of moles of CH3COOH in 250 mL of diluted vinegar = 0.003348 x (250/50) = 0.01674 mol No. of moles of CH3COOH in 25 mL of original vinegar = 0.01674 mol Molar mass of CH3COOH = 12x2 + 1x4 + 16x2 = 60 g mol-1 Mass of CH3COOH in 25 mL of original vinegar = mol x (molar mass) = 0.01674 x 60 = 1.004 g Acidity of the original vinegar = (1.004/25) x 100% = 4.016% 2008-06-15 11:09:44 補充： The 「acidity of the original vinegar」should be the「concentration of the original vinegar」instead.

其他解答:

aa.jpg

 

此文章來自奇摩知識+如有不便請留言告知

H+ + NaOH --> Na+ + H2O (1) NO. OF MOLE OF NaOH: 34.88 /1000 X 0.096=0.00334848 BY (1) No.of mole of vinegar = no .of mole of NaOH =0.00334848 No. of mole = Volume X acidity Acidity : 0.00334848/ (50/1000) =0.0669696 THE ORIGIN (NOT YEY DILUTE) VINEGAE ACIDIRY IS" 0.0669696/ (25/250) = 0.669696 =0.67 M
arrow
arrow
    文章標籤
    三角 文章 奇摩
    全站熱搜

    ohe09nz81r 發表在 痞客邦 留言(0) 人氣()

    E-mail轉寄