Titration (calculate Molarity)

[匿名]

2009-05-12 8:02 pm

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

回答 (1)

Uncle Michael

2009-05-15 9:31 pm

✔ 最佳答案

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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