Question

What is the molar solubility of pbbr2 in 0.500 m kbr solution?

Answer 1

Hey there!:

The Ksp of PbBr2 is 6.60*10⁻⁶

PbBr2 <=> Pb2+ + 2 Br-

Ksp = [Pb2+][Br-]2 = 6.60 x 10⁻⁶

Therefore:

In 0.500 M KBr solution:

[Br-] = [KBr] = 0.500 M assuming [Br-]PbBr2 << [Br-]KBr

Ksp = [Pb2+] * (0.500)² = 6.60 x 10⁻⁶

Molar solubility = [Pb²⁺] = 2.64 x 10⁻⁵ M

(Check [Br-]PbBr2 = 2 * ( 2.64 x 10⁻⁵M )  << 0.500 M so assumption is valid)

Answer : [Pb²⁺] = 2.64 x 10⁻⁵ M

Hope that helps!

Answer 2

The molar solubility of PbBr₂ in 0.500 KBr solution is 2.64 × 10⁻⁵ M

PbBr₂ is an ionic compound that dissociates in water to produce lead ions and Bromine ions.

The dissociation reaction can be expressed  as:

$\mathbf{PbBr_2 \rightleftharpoons Pb^{2+} + 2Br^-}$

The solubility product constant $\mathbf{Ksp}$ can be written as:

$\mathbf{K_{sp} = [Pb^{2+}] [Br^-]^2}$

If  $\mathbf{ [Pb^{2+}] = x \ ;}$ $\mathbf{[Br^-]= 2x}$

where;

• The Ksp for $\mathbf{PbBr_2 = 6.6 \times 10^{-6}}$

In a 0.500 kBr solution;

• KBr ≅ Br⁻

$\mathbf{6.6 \times 10^{-6 } = [x] [(2x)^2]}$

$\mathbf{6.6 \times 10^{-6 } = (x) \times (2x + 0.500)^2}$

Assuming that x = 0; since it is too small than 0.500

∴

$\mathbf{x = \dfrac{6.6 \times 10^{-6}}{0.500^2} }$

x = 2.64 × 10⁻⁵ M

Therefore, we can conclude that the molar solubility of PbBr₂ in 0.500 KBr solution is 2.64 × 10⁻⁵ M

Learn more about solubility product constant here:

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