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A saturated solution of Ag2CrO4 has a CrO42- concentration of 6.7 x 10-5 M. Calculate the molarity of Ag+ if the Ksp for Ag2CrO4 is equal to 1.2 x 10-12

Answer :

ChemistryHelper2024
To answer this question we must at the beginning write a balanced chemical equation:
Ag₂CrO₄ ↔  2 Ag⁺ + CrO₄²⁻

As we see from the equation that ionization of one mole of silver chromate gives two moles of silver and one mole of chromate 

So the solubility product Ksp = [Ag⁺]²[CrO₄⁻²]

[CrO₄⁻²] = 6.7 x 10⁻⁵ M and Ksp = 1.2 x 10⁻¹² (as given)
So:
1.2 x 10⁻¹² = [Ag⁺]² (6.7 x 10⁻⁵)

[Ag⁺]² = [tex] \frac{(1.2 x 10^{-12} )}{(6.7x10^{-5} ) } = 1.79 x 10^{-8} [/tex]

[tex][Ag^{+}] = \sqrt{1.79 x 10^{-8}} = 1.35 x 10^{-4} [/tex]

[Ag⁺] = 1.34 x 10⁻⁴ M 

Edufirst
Answer: 1.34 x 10⁻⁴M


Explanation:


1) Balanced ionic equation:


Ag₂CrO₄(s) ⇄ CrO₄²⁻ (aq)+ 2Ag²⁺ + (aq)

2) Ksp equation:


Ksp = [CrO₄²⁻] [Ag²⁺]²


3) Calculartions:


[Ag²⁺]² = Ksp / [CrO₄²⁻]

[Ag²⁺]² = 1.2x10⁻¹² / (6.7x10⁻⁵] = 1.79x10⁻⁸ M²


=> [Ag²⁺] = 1.34 x 10⁻⁴ M


Note: from the stoichiometry you get the same result. Since, the concentration of [Ag²⁺] is the double of the concentration of [CrO₄²⁻]:


2 x 6.7x10⁻⁵ M = 1.34 x 10⁻⁴ M

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