Given a 5.0 g sample of vinegar having a mass percentage of acetic acid equal to

Question

Given a 5.0 g sample of vinegar having a mass percentage of acetic acid equal to 9.1 (m/m) calculate what volume of NaOH solution of molarity 0.82 M will be necessary to titrate this vinegar sample. A) 8.96 mL B) 9.2 mL C) 9.245 mL D) 0.108 mL E) 9.0 mL Which of the following is true of a buffered solution? A) Any OH ions added will react with a weak acid already in solution. B) The solution will not its pH very much even if a concentrated acid is added. C) The solution will not change its pH very much even if a strong base is added. E) all of the above Which of following makes a buffer when added a solution of ammonium chloride? A) HCl B) NH_4Cl C) NH_3 D) CH_3COOH E) CH_3COONa

Explanation / Answer

(7)

Mass percentage of acetic acid sample = 9.1 %

It means that 100 g. of sample contains 91. g. of acetic acid

then, 5.0 g. of sample contains 5.0 * 9.1 / 100 = 0.455 g. of acetic acid

Moles of acetic acid = mass / molar mass = 0.455 / 60 = 0.00758 mol

CH3COOH (aq.) + NaOH (aq.) ------------> CH3COONa (aq.) + H2O (l)

SO, 1 mol of acetic acid needs 1 mol of NaOH

then, 0.00758 mol of acetic acid needs 0.00758 mol of NaOH

Therfore,

Volume of NaOH needed = 0.00758 * 1000 / 0.82 = 9.24 mL

So, (C)

(8)

(E) All the statements are correct regarding a buffer solution

(9)

(C) NH3

It is a basic buffer of weak base, NH3, and its salt with strong acid, NH4Cl

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