The voltage of the cell
Pt (s) I H2 ( 1 bar)I RNH2(0.1 M), RNH3+ Cl(0.05 M)IIS.H.E. is 0.49 V. Find kb of the base, RNH2
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To find the base dissociation constant, Kb, of RNH2, we can use the Nernst equation and the given cell voltage.
The Nernst equation relates the cell voltage (Ecell) to the standard cell potential (Eºcell), the gas constant (R), the temperature in Kelvin (T), the number of electrons transferred in the balanced cell equation (n), and the concentrations of the species involved:
Ecell = Eºcell - (RT/nF) * ln(Q)
Where:
Ecell = cell voltage (in volts)
Eºcell = standard cell potential (also in volts)
R = gas constant (8.314 J/(mol·K))
T = temperature (in Kelvin)
n = number of electrons transferred in the balanced cell equation
F = Faraday's constant (96485 C/mol)
Q = reaction quotient (the ratio of product concentrations to reactant concentrations)
In this case, the balanced cell equation is:
Pt(s) | H2(1 bar) | RNH2(0.1 M) || RNH3+ Cl(0.05 M) | Pt(s)
Since the cell is at equilibrium, the reaction quotient, Q, is equal to the equilibrium constant, K.
Therefore, the Nernst equation can be simplified to:
Ecell = Eºcell - (RT/nF) * ln(K)
We are given the cell voltage (Ecell) as 0.49 V. The standard cell potential (Eºcell) can be determined from tables or known values.
Assuming room temperature (25°C or 298 K), R = 8.314 J/(mol·K), and n = 2 (since two electrons are transferred in the balanced equation for this cell), we can substitute these values into the equation:
0.49 V = Eºcell - (8.314 J/(mol·K) * 298 K / (2 * 96485 C/mol)) * ln(K)
Simplifying further:
0.49 V = Eºcell - 0.0069 * ln(K)
Now, let's isolate ln(K) by rearranging the equation:
ln(K) = (Eºcell - 0.49 V) / 0.0069
Finally, we can determine K (the equilibrium constant):
K = e^(ln(K))
Once you have the value of K, you can use it to calculate Kb.
In this case, since KB is the base dissociation constant, it is related to the equilibrium constant using the equation:
KB = Kw / Ka (where Ka is the acid dissociation constant)
Therefore, once you have K, you can use it to calculate Ka. Then, with Ka and Kw (which is the ion product constant of water, 1x10^-14), you can calculate Kb using the above equation.