Concept explainers
Interpretation:
The value of pKw at the given temperatures has to be calculated. The standard enthalpy and entropy of autoprotolysis of water at 25 °C has to be determined.
Concept introduction:
At a given temperature, the product of the concentration of H+ and OH− is termed as the ionic product of the water at that temperature. The negative logarithm of ionic product gives pKw. The standard cell potential of a couple is the potential of the couple with respect to the potential of the standard hydrogen electrode. It is the potential of the cell under standard conditions of temperature, pressure and concentrations.
Answer to Problem 6.4IA
The value of pKw at the given temperatures is shown in the table below.
θ /°C | pKw |
20 °C | 14.22 |
25 °C | 14 |
30 °C | 11.82 |
The value of standard entropy is 9.6485 J K−1 mol−1_. The value of the standard enthalpy is −98625.5 J mol−1_.
Explanation of Solution
The cell is given below.
Pt|H2(g,pΘ)|NaOH(aq, 0.0100 mol kg−1),NaCl(aq, 0.01125 mol kg−1)|AgCl(s)|Ag(s)
The equilibrium constant for the hydrolysis of water Kw is expressed by the relation below.
Kw=[H2O][H+][OH−] (1)
Where,
- [H2O] is the concentration of water.
- [H+] is the concentration of H+.
- [OH−] is the concentration of OH−.
The value of pOH is calculated by the formula given below.
pOH=−log OH− (2)
The concentration of OH−, [OH−] is 0.0100 mol kg−1.
Substitute the value of [OH−] in equation (2).
pOH=−log (0.0100 )=2
The relation between pOH and pH is shown below.
pOH+pH=14 (3)
Substitute the value of pOH in equation (3).
2+pH=14pH=14−2=12 (4)
The Nernst equation is shown below
E=Eο−RTFln[H+]
Where,
- Ecell is the potential difference of the cell.
- EΘcell is the standard cell potential.
- R is the gas constant (8.314 J K-1 mol-1).
- T is the temperature.
- F is Faraday’s constant (96485 C mol−1).
Let the value of [H+] at 20 °C be x.
Let the value of [H+] at 25 °C be y.
Let the value of Ecell at 20 °C be E1.
Let the value of Ecell at 25 °C be E2.
Let the temperature 20 °C be T1.
Let the temperature 25 °C be T2.
Therefore,
E1=Eο−RT1Flnx (5)
And,
E2=Eο−RT2Flny (6)
Subtract equation (6) and (5).
T1lnx−T2lny=(E2−E1)FRT1logx−T2logy=(E2−E1)F2.303R (7)
The conversion of Celsius to Kelvin is done as,
0 οC=273 K
Therefore, the conversion of 20 °C to Kelvin is done as,
20 °C=20+273.15 K=293.15 K
The conversion of Celsius to Kelvin is done as,
0 οC=273 K
Therefore, the conversion of 25 °C to Kelvin is done as,
25 °C =25+273.15 K=298.15 K
The value of E1 is 1.04774 V.
The value of E2 is 1.04864 V.
The value of pH is calculated by the formula given below.
pH=−log H+
Therefore,
pH1=−log xlog x=−pH1
And,
pH2=−log ylog y=−pH2
Substitute the corresponding values in equation (7).
−(293.15 K×pH1)+(298.15 K×pH2)=(1.04864 V−1.04774 V)×96485 mol−12.303×8.314 J K-1 mol−1−(293.15 K×pH1)+(298.15 K×pH2)=0.0009 V ×96485 C mol−12.303×8.314 J K-1 mol−1×1 J1 CV−293.15 pH1+298.15 K pH2=4.53pH1=298.15 K pH2−4.53293.15
Substitute the value of pH from equation (4) in the above equation.
pH1=298.15 ×12−4.53293.15=12.19 (8)
The value of pKw is calculated by the formula given below.
pKw=pOH+pH (9)
Substitute the value of pH from equation (8) and pOH in equation (9).
pKw=2+12.19=14.19_
Therefore, the value of pKw at 20 °C is 14.19_.
Similarly, the value of pKw at various temperatures is calculated in the table below.
θ /°C | pKw |
20 °C | 14.19 |
25 °C | 14 |
30 °C | 11.82 |
The standard cell potential (EΘcell) is given by the Nernst equation as shown below.
EΘcell=Ecell+2.303RTFlogQ (10)
Where,
- E is the potential difference of the cell.
- R is the gas constant (8.314 J K-1 mol-1).
- T is the temperature.
- F is Faraday’s constant (96485 C mol−1).
- Q is the reaction quotient.
For the given reaction, the reaction quotient, Q can be calculated by the formula shown below.
Q=Concentration of NaClConcentration of NaOH (11)
The concentration of NaOH is 0.0100 mol kg−1.
The concentration of NaCl is 0.01125 mol kg−1.
Substitute the concentrations of NaOH and NaCl in equation (11).
Q=0.01125 mol kg−10.0100 mol kg−1=1.125
The value of temperature is 25 °C.
The conversion of Celsius to Kelvin is done as,
0 οC=273 K
Therefore, the conversion of 25 °C to Kelvin is done as,
25 °C =25+273.15 K=298.15 K
The value of Ecell at 25 °C is 1.04864 V.
Substitute the value of Ecell, Q, T, R and F in equation (11).
EΘcell=1.04864 V+2.303×8.314 J K-1 mol−1×298.15 K96485 C mol−1log1.125=1.04864 V+0.003026 J C−1×1 VC1 J=1.04864 V+0.003026 V=1.052 V
The value of temperature is 20 °C.
The conversion of Celsius to Kelvin is done as,
0 οC=273.15 K
Therefore, the conversion of 25 °C to Kelvin is done as,
20 °C =20+273.15 K=293.15 K
The value of Ecell at 20 °C is 1.04774 V.
Substitute the value of Ecell, Q, T, R and F in equation (11).
EΘcell=1.04774 V+2.303×8.314 J K-1 mol−1×293.15 K96485 C mol−1log1.125=1.04774 V+0.002975 J C−1×1 VC1 J=1.04774 V+0.002975 V=1.051 V
The value of temperature is 30 °C.
The conversion of Celsius to Kelvin is done as,
0 οC=273.15 K
Therefore, the conversion of 30 °C to Kelvin is done as,
30 °C =30+273.15 K=303.15 K
The value of Ecell at 30 °C is 1.04942 V.
Substitute the value of Ecell, Q, T, R and F in equation (11).
EΘcell=1.04942 V+2.303×8.314 J K-1 mol−1×303.15 K96485 C mol−1log1.125=1.04942 V+0.003077J C−1×1 VC1 J=1.04942 V+0.003077 V=1.053 V
The ΔEΘcell value is calculated below.
ΔEΘcell=1.052 V−1.051 V=0.001 V
The difference in temperature is calculated below.
ΔT=303.1 K−293.15 K=10 K
The value of (∂EΘcell∂T) is calculated below.
(∂EΘcell∂T)=0.001 V10 K=1.0×10−4 V K−1
The value of entropy (ΔSΘ) is calculated by the formula given below.
ΔSΘ=vF(∂EΘcell∂T) (12)
Where,
- ν is the number of electrons.
The number of electrons in the above cell is 1.
Substitute the corresponding values in equation (12).
ΔSΘ=1×96485 C mol−1×1.0 ×10−4 V K−1=9.6485 C V K−1 mol−1=9.6485 C V K−1 mol−1×1 J1 C V= 9.6485 J K−1 mol−1_
Therefore, the value of standard entropy is 9.6485 J K−1 mol−1_.
The standard Gibbs energy (ΔGΘ) is given by the expression,
ΔGΘ=−νFEΘcell (13)
The number of electrons in the given cell is 1.
The value of EΘcell is 1.052 V.
Substitute the value of ν, EΘcell, and F in equation (13).
ΔGΘ=−(1×96485 C mol−1×1.052 V)=−(101502.2 V C mol−1)= −(101502.2 V C mol−1×1 J1 V C)=−101502.2 J mol−1
The standard enthalpy (ΔHΘ) is calculated by the formula given below.
ΔHΘ=ΔGΘ+TΔSΘ (14)
The value of ΔSΘ is 9.6485 J K-1 mol-1
The value of ΔGΘ is −101502.2 J mol−1.
The value of temperature (T) is 298.15 K.
Substitute the value of ΔSΘ, ΔGΘ, and T in equation (14).
ΔHΘ=−101502.2 J mol−1+(298.15 K×9.6485 J K-1 mol-1)=−101502.2 J mol−1+2876.700 J mol−1=−98625.5 J mol−1_
Therefore, the value of the standard enthalpy is −98625.5 J mol−1_.
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