標題:
Titration (calculate Molarity)
Calculate the molarity of an acetic acid solution, HC2H3O2(aq), if a 25.0ml sample requires 27.2ml of 0.138 M NaOH in a titration. The following is my answer: (I would like to know, did I do it right?) (0.0272L) x (0.100 mole of base/ 1 L) x (2 mole of Acid/ 1 mold of Base) x (38.0g/ 1 mole of A) = 0.208g
最佳解答:
There are 3 mistakes in the answer: (1) The answer should be the molarity (molar concentration) of acetic acid, but not its mass. (2) The mole ratio of acid : base is equal to 1 : 1, but not 1 : 2. (3) The molar mass of the acid should be 60 g/mol, but not 38 g/mol. Actually, it is not necessary to use the molar mass of the acid in calculation. My answer is: CH3COOH + NaOH → CH3COONa + H2O Mole ratio CH3COOH : NaOH = 1 : 1 No. of moles of NaOH = 0.138 x 27.2 = MV = 0.00375 mol No. of moles of CH3COOH = No. of moles of NaOH = 0.003754 mol Molarity of CH3COOH = mol/V = 0.003754/(25/1000) = 0.150 M =
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Titration (calculate Molarity)
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發問:Calculate the molarity of an acetic acid solution, HC2H3O2(aq), if a 25.0ml sample requires 27.2ml of 0.138 M NaOH in a titration. The following is my answer: (I would like to know, did I do it right?) (0.0272L) x (0.100 mole of base/ 1 L) x (2 mole of Acid/ 1 mold of Base) x (38.0g/ 1 mole of A) = 0.208g
最佳解答:
There are 3 mistakes in the answer: (1) The answer should be the molarity (molar concentration) of acetic acid, but not its mass. (2) The mole ratio of acid : base is equal to 1 : 1, but not 1 : 2. (3) The molar mass of the acid should be 60 g/mol, but not 38 g/mol. Actually, it is not necessary to use the molar mass of the acid in calculation. My answer is: CH3COOH + NaOH → CH3COONa + H2O Mole ratio CH3COOH : NaOH = 1 : 1 No. of moles of NaOH = 0.138 x 27.2 = MV = 0.00375 mol No. of moles of CH3COOH = No. of moles of NaOH = 0.003754 mol Molarity of CH3COOH = mol/V = 0.003754/(25/1000) = 0.150 M =
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