The equilibrium constant expression for this reaction is:
Kc = ([Pb2+][H2])/([H+]^2)
At equilibrium, Kc = Qc, where Qc is the reaction quotient. Since the hydrogen gas pressure is maintained at 1.00 atm, the concentration of H2 is constant and can be represented as [H2] = Kp*P, where Kp is the equilibrium constant for the partial pressure of H2 and P is the pressure of H2 (1.00 atm). The equilibrium constant expression can then be rewritten as:
Kc = (Kp*P*[Pb2+])/([H+]^2)
Substituting the given values, we get:
0.025 = (Kp*1.00*[Pb2+])/([H+]^2)
Solving for [H+], we get:
[H+]^2 = (Kp*1.00*[Pb2+])/0.025
Taking the square root of both sides:
[H+] = sqrt((Kp*1.00*[Pb2+])/0.025)
To find the pH, we take the negative logarithm of [H+]:
pH = -log([H+])
Therefore, we can calculate the pH at which this reaction is at equilibrium by plugging in values for Kp and [Pb2+]. The value of Kp can be found from the equilibrium constant expression for the partial pressures:
Kp = (PH2/1)^2 = 1.00
Assuming the reaction takes place at standard conditions (25°C, 1 atm), we can look up the standard reduction potential for the half-reaction of Pb2+(aq) + 2e- -> Pb(s) in a table:
Pb2+(aq) + 2e- -> Pb(s) E° = -0.126 V
Using the Nernst equation, we can calculate the equilibrium constant expression for the reduction half-reaction, which is also the equilibrium constant expression for the oxidation half-reaction (since the overall reaction is a redox reaction). The Nernst equation is:
E = E° - (RT/nF)*lnQc
where E is the cell potential, R is the gas constant, T is the temperature in Kelvin, n is the number of electrons transferred (2 in this case), F is the Faraday constant, and Qc is the reaction quotient for the half-reaction. At equilibrium, E is zero, so the Nernst equation becomes:
0 = E° - (RT/nF)*lnKc
Solving for lnKc:
lnKc = E°/(RT/nF)
Substituting values:
lnKc = (-0.126 V)/(0.0257 V/K) = -4.899
Taking the exponential of both sides:
Kc = e^-4.899 = 1.38 x 10^-3
Now we can plug in values for Kp and [Pb2+] and calculate the pH at equilibrium:
pH = -log(sqrt((Kp*1.00*[Pb2+])/0.025))
pH = -log(sqrt((1.00*1.00*0.025)/(0.025*1.38 x 10^-3)))
pH = -log(sqrt(181.16))
pH = -log(13.46)
pH = 1.87
Therefore, the pH at which this reaction is at equilibrium is approximately 1.87.
Consider the reaction Pb(s) + 2H+(aq) —> Pb2+(aq) + H2(g). If the hydrogen gas pressure is maintained at 1.00 atm and the lead(II) ion concentration is 0.025 M, find the pH at which this reaction is at equilibrium.
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