This question is about hydrogen chloride and hydrochloric acid - AQA - GCSE Chemistry Combined Science - Question 6 - 2022 - Paper 1
Question 6
This question is about hydrogen chloride and hydrochloric acid.
1. Complete the dot and cross diagram to represent the bonding in hydrogen chloride on Figure 7.
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Worked Solution & Example Answer:This question is about hydrogen chloride and hydrochloric acid - AQA - GCSE Chemistry Combined Science - Question 6 - 2022 - Paper 1
Step 1
Complete the dot and cross diagram to represent the bonding in hydrogen chloride on Figure 7.
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Answer
The dot and cross diagram for hydrogen chloride (HCl) should display a shared pair of electrons between the hydrogen atom (H) and the chlorine atom (Cl). The chlorine atom contains six non-bonding electrons shown as dots (●), while the bonded pair is shown as a cross (×) overlapping to indicate the shared electrons. The final diagram should reflect this arrangement.
Step 2
What is meant by the term strong acid?
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A strong acid is one that completely ionizes in aqueous solution. This means that when the acid dissolves in water, it releases all its hydrogen ions (H⁺) into the solution, resulting in a high concentration of H⁺ ions and a low pH.
Step 3
Describe how magnesium can be used to distinguish between a strong acid and a weak acid of the same concentration.
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When magnesium is added to a strong acid, it reacts vigorously and fizzes, producing hydrogen gas quickly. In contrast, when added to a weak acid of the same concentration, the reaction occurs at a much slower rate and may produce less fizzing, indicating a weaker acid.
Step 4
What is the change in pH?
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The change in pH can be calculated by the formula: extpH=−extlog10[extH+]. When the concentration of hydrochloric acid is increased by a factor of 100, the resulting concentration of H⁺ ions also increases by 100 times. Hence, the change in pH is a decrease of 2 units, as each tenfold increase corresponds to a decrease of 1 in pH.
Step 5
Calculate the bond energy X.
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To calculate the bond energy X, we can apply the following steps:
Calculate the total bond energy of the bonds broken:
Bonds broken:
4×413+C=C+431=2083+C=C
Calculate the total bond energy of the bonds formed:
Bonds formed:
346+339+5×413=2750
The energy released in the reaction is given by
energy released=bonds broken−bonds formed
Thus,
56=2750−2083
which leads to
C=C=611 kJ/mol. Therefore, bond energy X is 611 kJ/mol.