Ag+ + Br- <-----> AgBr Equilibrium problem chemistry help?

[匿名]

2017-12-14 1:42 am

更新1:

Calculate the amount of AgBr solid that forms and the concentration of Ag+ and Br- ions remaining in solution if 0.500 L each of 1M solutions of AgNO3 and KBr are mixed together. Keq for the formation of AgBr from Ag+ and Br- is 1.87 x 10 ^12 Help me please...........apparently the answer is 0.5M of AgBr forms ,,, and the concentration of Ag+ and Br- is 7.3 x 10^-7 M how did they get this?

回答 (1)

戇戇居士

2017-12-14 2:39 am

After mixing and before reaction:
Initial amount of Ag⁺ = (1 mol/L) × (0.500 L) = 0.5 mol
Initial amount of Br⁻ = (1 mol/L) × (0.500 L) = 0.5 mol

Ag⁺(aq) + Br⁻(aq) → AgBr(s)
1 mole of Ag⁺ ion reacts with 1 mole of Br⁻ ion to give 1 mole AgBr precipitate.
Now, 0.5 mole of Ag⁺ ion and 0.5 mole of Br⁻ ion are used.
Hence, amount of AgBr precipitate formed = 0.5 mol

Consider the solubility equilibrium of AgBr :
____________ AgBr(s) __ ⇌ __ Ag⁺(aq) __ + __ Br⁻(aq) ___ Ksp = 1.87 × 10⁻¹²
Initial: ____________________ 0 M __________ 0 M
Change: __________________ +y M _________ +y M
At eqm: ___________________ y M __________ y M

Ksp = [Ag⁺] [Br⁻]
1.87 × 10⁻¹² = y²
y = √(1.87 × 10⁻¹²)
y = 1.37 × 10⁻⁶

Hence, [Ag⁺] = [Cl⁻] = 1.37 × 10⁻⁶ M

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