Solid Ba(NO3)2 is gradually dissolved in a 1.0 × 10-4 M Na2CO3 solution. At which concentration of Ba2+, precipitate of BaCO3 begins to form ? (Ksp for BaCO3 = 5.1 × 10-9)

Question

Solid Ba(NO3)2 is gradually dissolved in a 1.0 × 10-4 M Na2CO3 solution. At which concentration of Ba2+, precipitate of BaCO3 begins to form ? (Ksp for BaCO3 = 5.1 × 10-9) (A) 5.1 × 10-5 M (B) 7.1 × 10-8 M (C) 4.1 × 10-5 M (D) 8.1 × 10-7 M

Tardigrade Admin · Accepted Answer

Given Na2CO3 = 1.0 × 10-4M ∴ [CO3 --] = 1.0 × 10-4 M i.e. s=1.0×10-4M At equilibrium [Ba++] [CO3--] = Ksp of BaCO3 [Ba++] = (Ksp/ [CO3--]) = (5.1×10-9/1.0×10-4) = 5.1× 10-5 M

Q. Solid $B a (N O_{3})_{2}$ is gradually dissolved in a $1.0 \times 1 0^{- 4}$ M $N a_{2} C O_{3}$ solution. At which concentration of $B a^{2 +}$ , precipitate of $B a C O_{3}$ begins to form ? ( $K_{s p}$ for $B a C O_{3} = 5.1 \times 1 0^{- 9}$ )

$5.1 \times 1 0^{- 5} M$

$7.1 \times 1 0^{- 8} M$

$4.1 \times 1 0^{- 5} M$

$8.1 \times 1 0^{- 7} M$

Q. Solid Ba(NO3​)2​ is gradually dissolved in a 1.0×10−4 M Na2​CO3​ solution. At which concentration of Ba2+, precipitate of BaCO3​ begins to form ? (Ksp​ for BaCO3​=5.1×10−9)

5.1×10−5M

7.1×10−8M

4.1×10−5M

8.1×10−7M

Given Na2​CO3​=1.0×10−4M ∴[CO3−−​]=1.0×10−4M i.e.s=1.0×10−4M At equilibrium [Ba++][CO3−−​]=Ksp​ of BaCO3​ [Ba++]=[CO3−−​]Ksp​​=1.0×10−45.1×10−9​ =5.1×10−5M

Q. Solid $B a (N O_{3})_{2}$ is gradually dissolved in a $1.0 \times 1 0^{- 4}$ M $N a_{2} C O_{3}$ solution. At which concentration of $B a^{2 +}$ , precipitate of $B a C O_{3}$ begins to form ? ( $K_{s p}$ for $B a C O_{3} = 5.1 \times 1 0^{- 9}$ )

$5.1 \times 1 0^{- 5} M$

$7.1 \times 1 0^{- 8} M$

$4.1 \times 1 0^{- 5} M$

$8.1 \times 1 0^{- 7} M$