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Hydrochloric acid, HCl, is not a primary standard - Leaving Cert Chemistry - Question 2 - 2010

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Hydrochloric acid, HCl, is not a primary standard. A solution of hydrochloric acid, HCl, was standardised by titration with a 0.05 M solution of sodium carbonate, Na... show full transcript

Worked Solution & Example Answer:Hydrochloric acid, HCl, is not a primary standard - Leaving Cert Chemistry - Question 2 - 2010

Step 1

Name the pieces of apparatus A and B.

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Answer

A: Wash bottle
B: Volumetric flask

Step 2

Explain the terms (i) standard solution.

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Answer

A standard solution is a solution of known exact concentration, which serves as a reference in titrations. It is prepared using a substance that is stable, pure, and soluble in water.

Step 3

Explain the terms (ii) primary standard.

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Answer

A primary standard is a highly pure compound that can be used to determine the exact concentration of a solution. It must be stable, non-hygroscopic, and readily available. The primary standard is used to prepare a standard solution.

Step 4

What: Contains no water of crystallisation.

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Answer

The sodium carbonate used is anhydrous, meaning it does not contain water of crystallisation.

Step 5

Give: Stable / available pure / soluble in water.

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Answer

An example of a suitable primary standard is anhydrous sodium carbonate, which is stable, can be obtained in pure form, and is soluble in water.

Step 6

Describe: Rinsing the beaker (clock glass, glass rod) into the volumetric flask.

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Answer

To ensure all of the sodium carbonate solution is transferred, the glassware should be rinsed with the sodium carbonate solution into the volumetric flask.

Step 7

Describe: Topping up the volumetric flask.

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Answer

After transferring the solution, distilled water is added to the volumetric flask until it reaches the calibration mark on the neck to ensure the final volume is exactly 1 liter.

Step 8

Describe: Washing down the sides of the titration flask.

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Answer

When performing a titration, it is important to wash the sides of the flask with distilled water to ensure all reactants are in the solution and to prevent any loss of reagents.

Step 9

Show: 5.3 × 106 = 0.05.

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Answer

This demonstrates the calculation needed using the molar mass of sodium carbonate to find its concentration.

Step 10

Show: Clearly that a solution containing 5.3 g of anhydrous sodium carbonate, Na₂CO₃, in 1 litre is a 0.05 M solution.

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Answer

The molar mass of Na₂CO₃ is calculated as follows:

Na: 23 g/mol × 2 = 46 g/mol
C: 12 g/mol = 12 g/mol
O: 16 g/mol × 3 = 48 g/mol

Total: 46 + 12 + 48 = 106 g/mol

Now, using the equation:

ext{Molarity} = rac{ ext{moles of solute}}{ ext{liters of solution}}

We convert 5.3 g of Na₂CO₃ to moles:

ightarrow 0.050 moles$$ Thus, in 1 liter, the molarity is: $$ ext{M} = rac{0.050 ext{ moles}}{1 ext{ L}} = 0.05 M$$

Step 11

Find the molarity of the hydrochloric acid solution correct to two decimal places.

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Answer

To find the molarity (C) of the hydrochloric acid solution, we can use the titration data:

Using the formula from the titration reaction:

rac{27.5 ext{ cm}^3 imes M_1}{2} = rac{25.0 ext{ cm}^3 imes 0.05 ext{ M}}{1}

Solving for M1M_1:

M_1 = rac{25.0 imes 0.05 imes 2}{27.5}

Calculating the value gives:

M1=0.09extMM_1 = 0.09 ext{ M}

Therefore, the molarity of the hydrochloric acid solution is 0.09 M.

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