Question

At 25C, required volume of water, to dissolve 1g BaSo4 of Ksp 1.1×10^-10 will be???If molecular weight of BaSo4 is 233.​

Answer 1

Answer:

Mass of BaSO4 = 1 g

Molecular Weight of BaSO4 = 233 g/mol

Ksp = 1.1 * 10⁻¹⁰

According to the question, the dissociation reaction can be written as  

BaSo⁴ → Ba²⁺ + SO4²⁻

∴ Ksp = [Ba²⁺][SO4²⁻]

Let the solubility or the concentration of [Ba²⁺] & [SO4²⁻] be “s” mol/L each.

So,  

s * s = 1.1 * 10⁻¹⁰

⇒ s = √(1.1 * 10⁻¹⁰)  = 0.0000104 mol/L

Now, let's calculate,

The solubility in grams/L = 0.0000104 mol/L * 233 g/mol = 0.002423 g/L

Therefore,

To dissolve 1 g of BaSO4, the required amount of water will be  

= 1g / 0.002423 g/L

= 412.71 L

Hence, 412.71 L of water is required.

Answer 2

Answer:

Explanation:

The dissociation of BaSO₄ in water takes place as following:

BaSO₄↔ Ba²⁺ + SO₄²⁻

So its a salt of 1:1

then  

 Ksp=s¹⁺¹ [Ba²⁺][SO₄²⁻]

or

s²=Ksp  ;

s=$\sqrt{Ksp}$ ;

s=$\sqrt{1.1*10^{-10} }$ ;

s= 0.0000105M

Number of moles of  BaSO₄ in 1g= 1/233 moles

so the volume of water required to dissolve 1g of BaSO₄ is =$\frac{1}{233} * \frac{1}{0.0000105} =408.74 liters$