Question

The Ksp value for lead sulfide, PbS, is 1.0x10^-28. What is the concentration of Pb2+ in a saturated solution of lead sulfide?

a) 1.0x10^-56
b) 1.0x10^-14
C) 3.3X10^-29
d) 0.5x10^-14

Answer 1

Answer on Question #43057 - Chemistry - Other

Question:

The $K_{\mathrm{sp}}$ value for lead sulfide, PbS, is $1.0 \cdot 10^{-28}$. What is the concentration of $\mathrm{Pb}^{2+}$ in a saturated solution of lead sulfide?

a) $1.0 \cdot 10^{-56}$

b) $1.0 \cdot 10^{-14}$

c) $3.3 \cdot 10^{-29}$

d) $0.5 \cdot 10^{-14}$

Answer:

Lead sulfide dissociation equation is $\mathrm{PbS} \leftrightarrow \mathrm{Pb}^{2+} + \mathrm{S}^{2-}$. Thus, solubility product constant expression for PbS is $K_{\mathrm{sp}} = [\mathrm{Pb}^{2+}][\mathrm{S}^{2-}]$. As is clear from the dissociation equation, $[\mathrm{Pb}^{2+}] = [\mathrm{S}^{2-}]$. The solubility product constant expression for PbS may be written as $K_{\mathrm{sp}} = [\mathrm{Pb}^{2+}]^2$, whence

$$[Pb^{2+}] = \sqrt{K_{\mathrm{sp}}} = \sqrt{1.0 \cdot 10^{-28}} = 1.0 \cdot 10^{-14}$$

Correct answer is b) $1.0 \cdot 10^{-14}$

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