a) 0.132g of a pure carboxylic acid (R-COOH) was dissolved in 25.00mL of water and titrated with 0.120M of standardised NaOH solution.A volume of 14.80mL was required to reach the endpoint of the titration.Identify the carboxylic acid.b) A 0.670g sample of barium hydroxide is dissolved and diluted to the calibration mark in a 250.0mL volumetric flask. It was found that 11.56mL of this solution was needed to neutralise 25.00mL of nitric acid solution.Calculate the molarity of the nitric acid solution. Write a chemical equation in your response.c) One student's results are given below:- Concentration of NaOH(aq) = 0.110M- Volume of undiluted vinegar = 10.00mL- total volume of diluted vinegar = 100.00mL- volume of diluted vinegar used in each titration = 20.00mL- Avg. titre of NaOH(aq) = 15.35mLBased on these results, calculate the concentration, in mol L-1, of acetic acid in the undiluted vinegar solution.

Question
Asked Oct 30, 2019

a) 0.132g of a pure carboxylic acid (R-COOH) was dissolved in 25.00mL of water and titrated with 0.120M of standardised NaOH solution.

A volume of 14.80mL was required to reach the endpoint of the titration.
Identify the carboxylic acid.

b) A 0.670g sample of barium hydroxide is dissolved and diluted to the calibration mark in a 250.0mL volumetric flask. It was found that 11.56mL of this solution was needed to neutralise 25.00mL of nitric acid solution.

Calculate the molarity of the nitric acid solution. Write a chemical equation in your response.

c) One student's results are given below:
- Concentration of NaOH(aq) = 0.110M
- Volume of undiluted vinegar = 10.00mL
- total volume of diluted vinegar = 100.00mL
- volume of diluted vinegar used in each titration = 20.00mL
- Avg. titre of NaOH(aq) = 15.35mL
Based on these results, calculate the concentration, in mol L-1, of acetic acid in the undiluted vinegar solution.

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Expert Answer

Step 1

(a) The number of moles of NaOH (n) is calculated using equation (1) in which C is the concentration and V is the volume of solution.

п
с
V
. (1)
п
0.120 mol/L
0.0148 L
п%31.78x10-3 mol
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п с V . (1) п 0.120 mol/L 0.0148 L п%31.78x10-3 mol

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Step 2

One mole of NaOH neutralizes one mole of carboxylic acid. The number of moles of acid is equal to the number of moles of NaOH (n). The molar mass of acid (M) is calculated using equation (2) in which m is the mass of the acid dissolved.

т
(2)
п-
М
0.132 g
1.78 x10 mol
М
М-74.2 g/mol
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т (2) п- М 0.132 g 1.78 x10 mol М М-74.2 g/mol

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Step 3

The molar mass of the given acid is close to that of the prop...

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