Mercury(II) hydroxide is so insoluble in water that only about 2 formula units p

Question

Mercury(II) hydroxide is so insoluble in water that only about 2 formula units per billion dissolve:

Hg(OH)2(s) -> Hg2+(aq) + 2OH-(aq) Delta H = ?

It does, however, neutralize hydrochloric acid:

Hg(OH)2(s) + 2HCl(aq) -> HgCl2(aq) + 2H2O(l) Delta H = X

Using this chemical equation with a delta H value of X and the neutralization chemical equation in part B of the lab with an assumed H value of -54 kJ/mol, show how X and -54.0 kJ/mol can be combined to calculate the H for the dissolution of mercury(II) hydroxide in water.

Question options:

X - 54.0 kJ/mol = ?

2X - 54.0 kJ//mol = ?

X + 2(-54.0 kJ/mol) = ?

X - 2(-54.0 kJ/mol) = ?

2X - (-54.0 kJ/mol) = ?

-2X - 54.0 kJ/mol = ?

-X -(-54.0 kJ/mol) = ?

X - 54.0 kJ/mol = ?

2X - 54.0 kJ//mol = ?

X + 2(-54.0 kJ/mol) = ?

X - 2(-54.0 kJ/mol) = ?

2X - (-54.0 kJ/mol) = ?

-2X - 54.0 kJ/mol = ?

-X -(-54.0 kJ/mol) = ?

Explanation / Answer

in the given data,

dissolution of Hg(OH)2(s) requires energy. but in acid base neutralisation reaction energy releases.and also in the reaction given above, there are 2 mol of acid reacts with 2 mol of OH-.

so that net DH = X + 2(-54.0 kJ/mol)

answer: X + 2(-54.0 kJ/mol)

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