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Law of mass action and equilibrium constant

An aqueous solution contains an unknown concentration of {Ba}^

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Medium

Solving time: 3 mins

An aqueous solution contains an unknown concentration of $Ba^{2 +}$ . When $50 mL$ of a $1 M$ solution of $Na_{2} SO_{4}$ is added, $BaSO_{4}$ just begins to precipitate. The final volume is $500 mL$ . The solubility product of $BaSO_{4}$ is $1 \times 1 0^{- 10}$ . What is the original concentration of $Ba^{2 +}$ ?

5 \times 1 0^{- 9} M

2 \times 1 0^{- 9} M

1.1 \times 1 0^{- 9} M

1.0 \times 1 0^{- 10} M

Views: 5,332 students

Updated on: Dec 7, 2023

Book:

Arihant Chemistry JEE Main Chapterwise Solutions (2019-2002) (Arihant)

Exam:

JEE Mains 2018

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Evelyn Discussed

An aqueous solution contains an unknown concentration of

Ba^{2 +}

. When

50 mL

of a

1 M

solution of

Na_{2} SO_{4}

is added,

BaSO_{4}

just begins to precipitate. The final volume is

500 mL

. The solubility product of

BaSO_{4}

1 \times 1 0^{- 10}

. What is the original concentration of

Ba^{2 +}

11 mins ago

Discuss this question LIVE

11 mins ago

Text solutionVerified

Its given that the final volume is

500 mL

and this final volume was arrived when

50 mL

1 M

Na_{2} SO_{4}

was added to unknown

Ba^{2 +}

solution. So, we can interpret the volume of unknown

Ba^{2 +}

solution as

450 mL

i.e. as 450 mL i.e. 450mL + 50mL = 500mL

B a^{2}

N a_{2} S O_{4}

B a_{S} o_{4}

solution solution solution From this we can calculate the concentration of

SO_{4}^{2 -}

ion in the solution via

M_{1} V_{1} = M_{2} V_{2}

1 \times 50 = M_{2} \times 500

(as

1 M Na_{2} SO_{4}

is taken into consideration)

M_{2} = \frac{1}{10} = 0.1 M

Now for just precipitation, lonic product

=

Solubility product

(K_{sp})

i.e.

[Ba^{2 +}] [SO_{4}^{2 -}] = K_{sp}

BaSO_{4}

Given

K_{sp}

BaSO_{4} = 1 \times 1 0^{- 10}

So,

[Ba^{2 +}] [0.1] = 1 \times 1 0^{- 10}

[Ba^{2 +}] = 1 \times 1 0^{- 9} M

Remember This is the concentration of

Ba^{2 +}

ions in final solution. Hence, for calculating the

[Ba^{2 +}]

in original solution we have to use

M_{1} V_{1} = M_{2} V_{2}

as M_{1} \times 450 = 1 0^{- 9} \times 500

so, M_{1} = 1.1 \times 1 0^{- 9} M

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Question Text	An aqueous solution contains an unknown concentration of $Ba^{2 +}$ . When $50 mL$ of a $1 M$ solution of $Na_{2} SO_{4}$ is added, $BaSO_{4}$ just begins to precipitate. The final volume is $500 mL$ . The solubility product of $BaSO_{4}$ is $1 \times 1 0^{- 10}$ . What is the original concentration of $Ba^{2 +}$ ?
Updated On	Dec 7, 2023
Topic	Equilibrium
Subject	Chemistry
Class	Class 11
Answer Type	Text solution:1 Video solution: 5
Upvotes	566
Avg. Video Duration	8 min

Text solutionVerified

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An aqueous solution contains an unknown concentration of Ba2+. When 50 mL of a 1M solution of Na2​SO4​ is added, BaSO4​ just begins to precipitate. The final volume is 500 mL. The solubility product of BaSO4​ is 1×10−10. What is the original concentration of Ba2+ ?

Text solutionVerified

Filo tutor solutions (5)

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