Suppose the reaction Pb(s)+ 2H⁺(aq)$\rarr$Pb²⁺(aq)+ H₂(g)is carried out at pH = 4.00 and at a hydrogen gas pressure of 1.00 atm.The concentration of lead(II)ions that causes this reaction to be at equilibrium is

Question

Quizplus · Accepted Answer

First, we can write the expression for the equilibrium constant, K, using the law of mass action:
K = [PbS][H+][HS-]^2 / [Pb][H2S]^2

Next, we can set up an ICE (initial, change, equilibrium) table to track changes in concentration:
Pb(s)      + 2 H2S(aq)     <=> PbS(s)      + 2 H+(aq)    + 2HS-(aq)
I            C1                                     0                      0                         0
-2x        +x                                     +x                    +2x                    +x
E           C1-2x                                x                      2x                      x

We can substitute the expressions for the equilibrium concentrations into the equilibrium constant expression and solve for x:
K = [PbS][H+][HS-]^2 / [Pb][H2S]^2
K = (x)(2x)^2 / (C1-2x)(C1-2x)
Assuming x is small compared to initial concentrations, we can make the simplifying assumption that C1-2x = C1:
K = (x)(2x)^2 / (C1)^2
x = sqrt(K[C1]^2 / 8)
Note that K is given in the problem: K = 5.6 x 10^-29 at 25°C.
We also know that the initial concentration of H2S is equal to the solubility product constant at pH 4.0 (assuming negligible contribution from H+ ions), which we can calculate using the expression:
Ksp = [Pb2+][S2-] = 1.6 x 10^-28 at pH 4.0
Thus, [H2S] = sqrt(Ksp) = 1.3 x 10^-14 M.

Plugging in the given values, we obtain:
x = sqrt((5.6 x 10^-29)(1.3 x 10^-14)^2 / 8) = 2.5 x 10^-14 M
Therefore, the concentration of Pb2+ that causes the reaction to be at equilibrium is approximately 2.5 x 10^-14 M, which is equivalent to choice C.

Suppose the Reaction Pb(s)+ 2H+(aq) →\rarr→ Pb2+(aq)+ H2(g)is Carried Out at PH = 4

Suppose the Reaction Pb(s)+ 2H⁺(aq) $\rarr$ Pb²⁺(aq)+ H₂(g)is Carried Out at PH = 4