If 20 ml of 1·5 × 10 –5 M BaCl 2 solution is mixed with 40 ml of 0·9 × 10 –5 MNa 2 SO 4 solution, will a precipitate get formed? K sp for BaSO 4 = 0·1 × 10 –10 . - Wired Faculty

Question

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If 20 ml of 1·5 × 10^–5 M BaCl₂solution is mixed with 40 ml of 0·9 × 10^–5MNa₂SO₄ solution, will a precipitate get formed? K_sp for BaSO₄ = 0·1 × 10^–10.

Solution

Solubility product of BaSO₄ = 0·1 × 10^–10Total volume of the mixture solution = 20 + 40 = 60 ml
(i) Determining Ba²⁺ ions concentration ; (Before mixing) M₁ V₁ = M₂V₂ (After mixing)
$1.5 space cross times space 10 to the power of negative 5 end exponent space cross times space 20 space equals space straight M subscript 2 space cross times space 60 therefore space space space space straight M subscript 2 left parenthesis Molarity space of space BaCl subscript 2 right parenthesis space space space space equals space fraction numerator 1.5 space cross times space 10 to the power of negative 5 end exponent space cross times space 20 over denominator 60 end fraction equals space 0.5 space cross times space 10 to the power of negative 5 end exponent space mol space straight L to the power of negative 1 end exponent therefore space space space space open square brackets Ba to the power of 2 plus end exponent close square brackets space space equals space 0.5 space cross times space 10 to the power of negative 5 end exponent space mol space ions space straight L to the power of minus to the power of 1$
$left parenthesis ii right parenthesis space Determining space SO subscript 4 superscript 2 minus end superscript space ions space concentration colon left parenthesis Before space mixing right parenthesis space straight M subscript 1 straight V subscript 1 space equals space straight M subscript 2 straight V subscript 2 left parenthesis After space mixing right parenthesis space space space space space space space 0.9 space cross times space 10 to the power of negative 5 end exponent space cross times space 40 space equals space straight M subscript 2 space cross times space 60 therefore space space space space straight M subscript 2 left parenthesis Molarity space of space Na subscript 2 SO subscript 4 right parenthesis space space equals space fraction numerator 0.9 space cross times space 10 to the power of negative 5 end exponent space cross times space 40 over denominator 60 end fraction equals space 0.6 space cross times space 10 to the power of negative 5 end exponent space mol space straight L to the power of negative 1 end exponent$
As sodium sulphate is fully ionised
$space space Na subscript 2 SO subscript 4 space space space rightwards harpoon over leftwards harpoon space space space space 2 Na to the power of 2 plus end exponent space plus space SO subscript 4 superscript 2 minus end superscript therefore space space open square brackets SO subscript 4 superscript negative 2 end superscript close square brackets space equals space 0.6 space cross times space 10 to the power of negative 5 end exponent space mol space ions space straight L to the power of negative 1 end exponent$
(iii) Determining ionic product;
$open square brackets Ba to the power of 2 plus end exponent close square brackets space open square brackets SO subscript 4 superscript 2 minus end superscript close square brackets space equals space left parenthesis 0.5 space cross times space 10 to the power of negative 5 end exponent right parenthesis thin space left parenthesis 0.6 space cross times space 10 to the power of negative 5 end exponent right parenthesis space space space space space space space space space space space space space space space space space space space space space space equals space 0.30 space cross times space 10 to the power of negative 10 end exponent$
As the ionic product of BaSO₄ (0·30 × 10^–10) is more than the K_sp value of the salt, therefore, a precipitate of barium sulphate will be formed.

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Equilibrium

If 20 ml of 1·5 × 10^–5 M BaCl₂solution is mixed with 40 ml of 0·9 × 10^–5MNa₂SO₄ solution, will a precipitate get formed? K_sp for BaSO₄ = 0·1 × 10^–10.

Some More Questions From Equilibrium Chapter

What is equilibrium?

Give two examples of physical equilibrium.

What kind of molecules in a liquid can evaporate?

Name the factors on which vapour pressure of any liquid depends.

Which measurable property becomes constant in water vapour equilibrium?

Give an example from daily life in which there is gas solution equilibrium?

A crystal of common salt of a given mass is kept in its aqueous solution. After 24 hours, its mass remains the same. Is the crystal in equilibrium with the solution?

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Equilibrium

If 20 ml of 1·5 × 10–5 M BaCl2 solution is mixed with 40 ml of 0·9 × 10–5MNa2SO4 solution, will a precipitate get formed? Ksp for BaSO4 = 0·1 × 10–10.

Some More Questions From Equilibrium Chapter

What is equilibrium?

Give two examples of physical equilibrium.

What kind of molecules in a liquid can evaporate?

Name the factors on which vapour pressure of any liquid depends.

Which measurable property becomes constant in water vapour equilibrium?

Give an example from daily life in which there is gas solution equilibrium?

A crystal of common salt of a given mass is kept in its aqueous solution. After 24 hours, its mass remains the same. Is the crystal in equilibrium with the solution?

Mock Test Series

NCERT Sample Papers

Entrance Exams Preparation

If 20 ml of 1·5 × 10^–5 M BaCl₂solution is mixed with 40 ml of 0·9 × 10^–5MNa₂SO₄ solution, will a precipitate get formed? K_sp for BaSO₄ = 0·1 × 10^–10.