SSCE VCE Chemistry Acid-Base Reactions: 0.132 g of a pure carboxylic aci

0.132 g of a pure carboxylic acid (R-COOH) was dissolved in 25.00 mL of water and titrated with 0.120 M NaOH solution - VCE - SSCE Chemistry - Question 8 - 2009 - Paper 1

Question 8

0.132 g of a pure carboxylic acid (R-COOH) was dissolved in 25.00 mL of water and titrated with 0.120 M NaOH solution. A volume of 14.80 mL was required to reach the... show full transcript

Worked Solution & Example Answer:0.132 g of a pure carboxylic acid (R-COOH) was dissolved in 25.00 mL of water and titrated with 0.120 M NaOH solution - VCE - SSCE Chemistry - Question 8 - 2009 - Paper 1

Step 1

Determine the moles of NaOH used

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Answer

First, calculate the moles of NaOH used in the titration using the formula:

ext{moles} = ext{concentration} imes ext{volume}

Given:

Concentration of NaOH = 0.120 M
Volume of NaOH used = 14.80 mL = 0.01480 L

Thus,

ext{moles NaOH} = 0.120 imes 0.01480 = 0.001776 ext{ moles}

Step 2

Determine moles of the carboxylic acid

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Answer

Since the reaction between the carboxylic acid and NaOH is a 1:1 mole ratio, the moles of the carboxylic acid will also be:

ext{moles of R-COOH} = 0.001776 ext{ moles}

Step 3

Calculate the molar mass of the carboxylic acid

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Answer

Now, calculate the molar mass (MM) of the carboxylic acid using its mass and moles:

ext{MM} = rac{ ext{mass}}{ ext{moles}} = rac{0.132 ext{ g}}{0.001776 ext{ moles}} \ = 74.3 ext{ g/mol}

Step 4

Identify the correct carboxylic acid

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Answer

Now, compare the calculated molar mass (74.3 g/mol) with the given options:

A. HCOOH (formic acid): MM = 46.03 g/mol
B. CH3COOH (acetic acid): MM = 60.05 g/mol
C. C2H4COOH (propanoic acid): MM = 74.08 g/mol
D. C3H6COOH (butanoic acid): MM = 74.10 g/mol

The closest match is option D (C3H6COOH), which falls within the expected range for errors. Thus, the answer is D.

0.132 g of a pure carboxylic acid (R-COOH) was dissolved in 25.00 mL of water and titrated with 0.120 M NaOH solution - VCE - SSCE Chemistry - Question 8 - 2009 - Paper 1

Worked Solution & Example Answer:0.132 g of a pure carboxylic acid (R-COOH) was dissolved in 25.00 mL of water and titrated with 0.120 M NaOH solution - VCE - SSCE Chemistry - Question 8 - 2009 - Paper 1

Determine the moles of NaOH used

Determine moles of the carboxylic acid

Calculate the molar mass of the carboxylic acid

Identify the correct carboxylic acid

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