When solid lead iodide is added to water, the equilibrium concentration of I- becomes 2.6× 10- 3 M. What is the Ks p for PbI2 ?

Question

When solid lead iodide is added to water, the equilibrium concentration of I- becomes 2.6× 10- 3 M. What is the Ks p for PbI2 ? (A) 2.2× 10- 9 (B) 8.8× 10- 9 (C) 1.8× 10- 8 (D) 3.5× 10- 8

Tardigrade Admin · Accepted Answer

The equilibrium is PbI2leftharpoons Pb2 ++2I- On the basis of this equation, the concentration of Pb2 + ions will be half of the concentration of I- ions. Thus, [.I-].=2.6× 10- 3M and [.Pb2 +].=1.3× 10- 3M Ks p=[.Pb2 +].[.I-].2=[.1.3× 10- 3M].[.2.6× 10- 3M].2 Ks p=8.8× 10- 9.

Q. When solid lead iodide is added to water, the equilibrium concentration of $I^{-}$ becomes $2.6 \times 1 0^{- 3}$ M. What is the $K_{s p}$ for $P b I_{2}$ ?

$2.2 \times 1 0^{- 9}$

$8.8 \times 1 0^{- 9}$

$1.8 \times 1 0^{- 8}$

$3.5 \times 1 0^{- 8}$

Q. When solid lead iodide is added to water, the equilibrium concentration of I− becomes 2.6×10−3 M. What is the Ksp​ for PbI2​ ?

2.2×10−9

8.8×10−9

1.8×10−8

3.5×10−8

The equilibrium is PbI2​⇌Pb2++2I− On the basis of this equation, the concentration of Pb2+ ions will be half of the concentration of I− ions. Thus, [I−]=2.6×10−3M and [Pb2+]=1.3×10−3M Ksp​=[Pb2+][I−]2=[1.3×10−3M][2.6×10−3M]2 Ksp​=8.8×10−9.

Q. When solid lead iodide is added to water, the equilibrium concentration of $I^{-}$ becomes $2.6 \times 1 0^{- 3}$ M. What is the $K_{s p}$ for $P b I_{2}$ ?

$2.2 \times 1 0^{- 9}$

$8.8 \times 1 0^{- 9}$

$1.8 \times 1 0^{- 8}$

$3.5 \times 1 0^{- 8}$