Chemistry, asked by kanishgoyal, 10 months ago

solubility of baso4 in pure water is 4 into 10 power minus 5 m solubility of baso4 in 0.01 m bacl2 solution will be​

Answers

Answered by abhi178
7

solubility of barium sulphate in 0.01M of barium chloride is 1.6 × 10^-7

It has given that solubility of BaSO₄ in pure water is 4 × 10¯5 .

dissociation reaction of BaSO₄ is...

BaSO₄ ⇔Ba²⁺ + SO₄²¯

solubility constant, Ksp = [Ba²⁺][SO₄²¯]

= (4 × 10^-5)(4 × 10^-5)

= 1.6 × 10^-9

let s be the solubility of BaSO₄ in 0.01M BaCl₂ solution.

[Ba²⁺] = 0.01 + y ≈ y [ because y << 0.01 ]

[SO₄²¯] = 0.01

now solubility constant, Ksp = [Ba²⁺][SO₄¯]

⇒1.6 × 10^-9 = y × 0.01

⇒1.6 × 10^-9 = 0.01y

⇒y = 1.6 × 10^-7

therefore solubility of barium sulphate in 0.01M of barium chloride is 1.6 × 10^-7

Answered by jenishanto2004
3

Answer:

Explanatsolubility of barium sulphate in 0.01M of barium chloride is 1.6 × 10^-7

It has given that solubility of BaSO₄ in pure water is 4 × 10¯5 .

dissociation reaction of BaSO₄ is...

BaSO₄ ⇔Ba²⁺ + SO₄²¯

solubility constant, Ksp = [Ba²⁺][SO₄²¯]

= (4 × 10^-5)(4 × 10^-5)

= 1.6 × 10^-9

let s be the solubility of BaSO₄ in 0.01M BaCl₂ solution.

[Ba²⁺] = 0.01 + y ≈ y [ because y << 0.01 ]

[SO₄²¯] = 0.01

now solubility constant, Ksp = [Ba²⁺][SO₄¯]

⇒1.6 × 10^-9 = y × 0.01

⇒1.6 × 10^-9 = 0.01y

⇒y = 1.6 × 10^-7

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