Question

At 25.0∘c, the molar solubility of silver chromate in water is 1.10×10−12 m . calculate the solubility in grams per liter. express your answer in grams per liter to three significant figures. how many liters of water are required to dissolve 1.00 g of silver chromate? express your answer in liters to three significant figures.

Answer 1

Answer: Solubility is $3.65\times 10^{-10}grams/liter$ and $2.74\times 10^{9}L$ of water is required to dissolve 1 g of silver chromate.

Explanation: The equation for the reaction will be as follows:

$Ag_2CrO_4\leftrightharpoons 2Ag^++CrO_4^{2-}$

1 mole of $Ag_2CrO_4$ gives 2 moles of $Ag^{+}$ and 1 mole of $CrO_4^{2-}$

Thus if solubility of $Ag_2CrO_4$ is s moles/liter, solubility of $Ag^{+}$ is 2s moles\liter and solubility of $CrO_4^{2-}$ is s moles/liter

Given:- $s=1.10\times 10^{-4}moles/liter$

Solubility in grams/liter=$\text{solubility in moles/liter}\times {\text {Molar mass}}$

Solubility in grams/liter=$1.10\times 10^{-12}\times 331.73=3.65\times 10^{-10}grams/liter$

$3.65\times 10^{-10}g$ of $Ag_2CrO_4$ dissolve in= 1L

1 g of $Ag_2CrO_4$ dissolve in=$\frac{1}{3.65\times 10^{-10}}\times 1=2.74\times 10^{9}L$

Answer 2

The solubility silver chromate in grams per liter = 3.65. 10⁻¹⁰

2,74.10⁹ liters of water are required to dissolve 1.00 g of silver chromate

Further explanation

Solubility is the maximum amount of a substance that can dissolve in some solvents. Factors that affect solubility

• 1. Temperature:

• 2. Surface area:

• 3. Solvent type:

• 4. Stirring process:

Ksp is an ion product in equilibrium

Solubility relationships and solubility constants of the AxBa solution can be stated as follows.

AₓBₐ (s) ← ⎯⎯⎯⎯ → x Aᵃ⁺ (aq) + a Bˣ⁻ (aq)

s                                xs              ys

Ksp = [Aᵃ⁺] ˣ [Bˣ⁻] ᵃ

Ksp = (xs) ˣ (ys) ᵃ

Solubility unit in the form of mole / liter or gram / liter

Solubility of Ag₂CrO₄ = 1.10 .10⁻¹² mole / liter

to change units to grams / liter, we multiply by molar mass:

M Ag₂CrO₄ = 2.Ar Ag + Ar Cr + 4. Ar O

M Ag₂CrO₄ = 2.108 + 52 + 4.16

M Ag₂CrO₄ = 332

So the solubility is in grams / liter

= 1.10. 10⁻¹² x 332

= 3.65. 10⁻¹⁰

(3 significant numbers, 3.6 and 5)

If we dissolve 1 gram of silver chromate into the solution, we need as much water as possible

= 1 gram / 3.65. 10⁻¹ grams / liter

= 2.74.10⁹ liter

(3 significant numbers, 2,7 and 4)

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Keywords: solubility, silver chromate, a significant number