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soldi70 [24.7K]
1 year ago
13

5. Citric acid is the main acid present in lemon juice. However vitamin C, or ascorbic acid is

Chemistry
1 answer:
aev [14]1 year ago
7 0

In the titration of lemon juice, the presence of ascorbic acid means the concentration of citric acid you calculated is higher.

An acid-base titration is a common way to determine the unknown concentration of an acid, given we know the concentration of the base and determine the spent volume in the titration. Let's consider the neutralization reactions that take place in a mixture of citric acid and ascorbic acid.

Citric acid titration :

3 NaOH(aq)  +  H₃C₆H₅O₇(aq)  →  Na₃C₆H₅O₇(aq)  +  3 H₂O(l)

Ascorbic acid titration:

NaOH(aq) + HC₆H₇O₆(aq)  →   NaC₆H₇O₆(aq)   +  H₂O(l)

If we titrated a solution that contained only citric acid, we can relate through stoichiometry the moles and concentration of citric acid. However, if the solution also contained ascorbic acid, we would have to spend more NaOH to titrate it. Since more NaOH would react, we would conclude that there is more citric acid to react, calculating a higher concentration of the same.

In the titration of lemon juice, the presence of ascorbic acid means the concentration of citric acid you calculated is higher.

You can learn more about titration here: brainly.com/question/2728613

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A sample of an unknown metal has a mass of 58.932g. it has been heated to 101.00 degrees C, then dropped quickly into 45.20 mL o
yaroslaw [1]
<h3>Answer:</h3>

0.111 J/g°C

<h3>Explanation:</h3>

We are given;

  • Mass of the unknown metal sample as 58.932 g
  • Initial temperature of the metal sample as 101°C
  • Final temperature of metal is 23.68 °C
  • Volume of pure water = 45.2 mL

But, density of pure water = 1 g/mL

  • Therefore; mass of pure water is 45.2 g
  • Initial temperature of water = 21°C
  • Final temperature of water is 23.68 °C
  • Specific heat capacity of water = 4.184 J/g°C

We are required to determine the specific heat of the metal;

<h3>Step 1: Calculate the amount of heat gained by pure water</h3>

Q = m × c × ΔT

For water, ΔT = 23.68 °C - 21° C

                       = 2.68 °C

Thus;

Q = 45.2 g × 4.184 J/g°C × 2.68°C

    = 506.833 Joules

<h3>Step 2: Heat released by the unknown metal sample</h3>

We know that, Q =  m × c × ΔT

For the unknown metal, ΔT = 101° C - 23.68 °C

                                              = 77.32°C

Assuming the specific heat capacity of the unknown metal is c

Then;

Q = 58.932 g × c × 77.32°C

   = 4556.62c Joules

<h3>Step 3: Calculate the specific heat capacity of the unknown metal sample</h3>
  • We know that, the heat released by the unknown metal sample is equal to the heat gained by the water.
  • Therefore;

4556.62c Joules = 506.833 Joules

c = 506.833 ÷4556.62

  = 0.111 J/g°C

Thus, the specific heat capacity of the unknown metal is 0.111 J/g°C

8 0
3 years ago
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