An aqueous solution contains an un known concentration of $B a^{2 +}$ When $50 m L$ of a 1 M solution of $N a_{2} S O_{4}$ is added, $B a S O_{4}$ just begins to precipitate. The final volume is 500 mL. The solubility product of $B a S O_{4}$ is $1 \times 10^{- 10}$ .
What is the original concentration of $B a^{2 +}$ .

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

$5 \times 10^{- 9} M$

$2 \times 10^{- 9} M$

$1.1 \times 10^{- 9} M$

$1.0 \times 10^{- 10} M$

answer is C.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

Solution: Final concentration of $[S O_{4}^{-}] = \frac{50 \times 1}{500} = 0.1 M$
$K_{s p} o f B_{a} S O_{4} [B a^{2 +}] [S O_{4}^{2 -}] = 1 \times 10^{- 10} [B a^{2 +}] [0.1] = 1 \times 10^{- 10} B a^{2 +} = \frac{10^{- 10}}{0.1} = 10^{- 9} M$
Concentration of $B a^{2 +}$ in final solution $= 10^{- 9} M$
Concentration of $B a^{2 +}$ in original
In solution, $M_{1} . V_{1} = M_{2} V_{2}$
$M_{1} (500 - 50) = 10^{- 9} (500)$
$M_{1} = \frac{500 \times 10^{- 9}}{450}$
$M_{1} = 1.11 \times 10^{- 9} M$