What is the maximum concentration of equimolar solutions of FeSO 4 and Na 2 S so that when mixed in equal volumes, there is no precipitation of FeS ? ( K sp . of FeS .=6.25 × 10-18)

Question

What is the maximum concentration of equimolar solutions of FeSO 4 and Na 2 S so that when mixed in equal volumes, there is no precipitation of FeS ? ( K sp . of FeS .=6.25 × 10-18) (A) 5 × 10-9 M (B) 5 × 10-8 M (C) 5 × 10-18 M (D) 5 × 109 M

Tardigrade Admin · Accepted Answer

s 2=6.25 × 10-18 s =2.5 × 10-9 M =2 s=2 × 2.5 × 10-9 =5.0 × 10-9 M

Q. What is the maximum concentration of equimolar solutions of $F e S O_{4}$ and $N a_{2} S$ so that when mixed in equal volumes, there is no precipitation of $F e S$ ? $(K_{s p}$ of $F e S$ $= 6.25 \times 1 0^{- 18})$

$5 \times 1 0^{- 9} M$

$5 \times 1 0^{- 8} M$

$5 \times 1 0^{- 18} M$

$5 \times 1 0^{- 9} M$

$s^{2} = 6.25 \times 1 0^{- 18}$
$s = 2.5 \times 1 0^{- 9}$
$M = 2 s = 2 \times 2.5 \times 1 0^{- 9}$
$= 5.0 \times 1 0^{- 9} M$

Q. What is the maximum concentration of equimolar solutions of FeSO4​ and Na2​S so that when mixed in equal volumes, there is no precipitation of FeS ? (Ksp​ of FeS =6.25×10−18)

5×10−9M

5×10−8M

5×10−18M

5×10−9M

s2=6.25×10−18 s=2.5×10−9 M=2s=2×2.5×10−9 =5.0×10−9M

Q. What is the maximum concentration of equimolar solutions of $F e S O_{4}$ and $N a_{2} S$ so that when mixed in equal volumes, there is no precipitation of $F e S$ ? $(K_{s p}$ of $F e S$ $= 6.25 \times 1 0^{- 18})$

$5 \times 1 0^{- 9} M$

$5 \times 1 0^{- 8} M$

$5 \times 1 0^{- 18} M$

$5 \times 1 0^{- 9} M$

$s^{2} = 6.25 \times 1 0^{- 18}$
$s = 2.5 \times 1 0^{- 9}$
$M = 2 s = 2 \times 2.5 \times 1 0^{- 9}$
$= 5.0 \times 1 0^{- 9} M$