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.
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.
To find the pH at which the reaction is at equilibrium, we need to determine the concentration of H+ ions at equilibrium.
Given:
Pressure of H2 gas (PH2) = 1.00 atm
Concentration of Pb2+ ions (Pb2+) = 0.025 M
The reaction is:
Pb(s) + 2H+(aq) → Pb2+(aq) + H2(g)
First, we need to determine the concentration of H+ ions. Since the reaction equation shows that 2 moles of H+ ions are consumed for every 1 mole of H2 produced, the concentration of H+ ions is half the concentration of Pb2+ ions.
[H+] = 0.025 M / 2
[H+] = 0.0125 M
Now, we can calculate the pH using the following equation:
pH = -log[H+]
pH = -log(0.0125)
pH ≈ 1.90
Therefore, at equilibrium, the pH of the system is approximately 1.90.
To find the pH at which this reaction is at equilibrium, we first need to understand the relationship between pH and hydrogen ion concentration.
The pH scale is a logarithmic scale that measures the acidity or alkalinity of a solution. It is defined as the negative logarithm (base 10) of the hydrogen ion concentration ([H+]). Mathematically, it can be represented as:
pH = -log10[H+]
In this case, since the hydrogen gas pressure is maintained at 1.00 atm, we know that the concentration of H+ ions produced from the reaction is also 1.00 M. Note that the concentration of H+ ions in a solution is proportional to the hydrogen gas pressure.
Now, let's analyze the given reaction:
Pb(s) + 2H+(aq) —> Pb2+(aq) + H2(g)
According to the reaction stoichiometry, every 2 moles of H+ ions consumed produce 1 mole of H2 gas. Therefore, at equilibrium, the concentration of H+ ions will be halved (0.5 M), while the concentration of Pb2+ ions will be equal to the initial concentration (0.025 M).
To calculate the pH at equilibrium, we can use the formula:
pH = -log10[H+]
Substituting [H+] = 0.5 M into the equation, we get:
pH = -log10(0.5) ≈ 0.301
Therefore, at equilibrium, the pH of the solution will be approximately 0.301.