Experimentally, the self-dissociation constants of $\ce{H2O}$ to $\ce{H3O+}$ and $\ce{OH-}$ and $\ce{D2O}$ to $\ce{D3O+}$ and $\ce{OD-}$ can be measured as in The Ionization Constant of Deuterium Oxide from 5 to 50 [degrees] J. Phys. Chem., 1966, 70, pp 3820–3824 and it is found that $\ce{H2O}$ is about 8 times more dissociated (equilibrium constant is 8 times greater).
But using the above data to say $\ce{D3O+}$ is stronger is misleading, because this corresponds to a reaction with $\ce{OD-}$, not $\ce{D2O}$. $\ce{D3O+}$ simply has a lower concentration in heavy water than $\ce{H3O+}$ has in light water.
As for why the $\ce{D2O}$ is less dissociated than $\ce{H2O}$, The ionization constant of heavy water ($\ce{D2O}$) in the temperature range 298 to 523 K Canadian Journal of Chemistry, 1976, 54(22): 3553-3558 breaks the differences down in to enthalpy and entropy components, which both favor ionization of $\ce{H2O}$ and states that $\ce{D2O}$ is a more structured liquid than $\ce{H2O}$. Not only the bonds of each product and reactant molecule need to be considered, but also the intermolecular forces: the number and strength of intermolecular hydrogen bonds for each species. See Quantum Differences between Heavy and Light Water Physical Review Letters 101, 065502 for recent (2008) experimental data.
Numerous references characterized $\ce{D2O}$ as "more structured" than $\ce{H2O}$, meaning more hydrogen bonds, and a more narrow distribution of hydrogen bond lengths and angles. According to Effect of Ions on the Structure of Water: Structure Making and Breaking Chem. Rev. 2009, 109, 1346–1370 "It is indeed generally agreed that heavy water, $\ce{D2O}$, is more strongly hydrogen bonded (structured) than light water, $\ce{H2O}$." My explanation would therefore be that there is a greater penalty for placing ions in $\ce{D2O}$ than $\ce{H2O}$ as far as disruption of a hydrogen bonding network.
Explanation of Normal/Inverse Solvent Isotope Effect
For a kinetic normal/inverse solvent isotope effect
there will be a reactant and transition state. If (for example) there is a single solvent exchangeable proton that is the same group in the reactant and transition state, for example, $\ce{ROH}$ in the reactant and $\ce{R'OH}$ in the transition state, switching solvents from $\ce{H2O}$ to $\ce{D2O}$ will either favor the reactant or the transition state relative to each other (considering the respective $\ce{OH}$ bond strengths as well and intermolecular hydrogen bonds to solvent). IF $\ce{D2O}$ favors the reactant relative to the transition state (activation energy is increased), this is a "normal kinetic solvent isotope effect". Oppositely, if $\ce{D2O}$ favors the transition state relative to the reactant this is an "inverse kinetic solvent isotope effect." More complex scenarios involving more exchangeable sites can of course occur.
Similarly there can be equilibrium normal/inverse solvent isotope effect, if there is an equilibrium reaction and then it is reactant vs. product (rather than reactant vs. transition state) that matters.
Experimentally, the self-dissociation constants of $\ce{H2O}$ to $\ce{H3O+}$ and $\ce{OH-}$ and $\ce{D2O}$ to $\ce{D3O+}$ and $\ce{OD-}$ can be measured as in The Ionization Constant of Deuterium Oxide from 5 to 50 [degrees] J. Phys. Chem., 1966, 70, pp 3820–3824 and it is found that $\ce{H2O}$ is about 8 times more dissociated (equilibrium constant is 8 times greater).
But using the above data to say $\ce{D3O+}$ is stronger is misleading, because this corresponds to a reaction with $\ce{OD-}$, not $\ce{D2O}$. $\ce{D3O+}$ simply has a lower concentration in heavy water than $\ce{H3O+}$ has in light water.
As for why the $\ce{D2O}$ is less dissociated than $\ce{H2O}$, The ionization constant of heavy water ($\ce{D2O}$) in the temperature range 298 to 523 K Canadian Journal of Chemistry, 1976, 54(22): 3553-3558 breaks the differences down in to enthalpy and entropy components, which both favor ionization of $\ce{H2O}$ and states that $\ce{D2O}$ is a more structured liquid than $\ce{H2O}$. Not only the bonds of each product and reactant molecule need to be considered, but also the intermolecular forces: the number and strength of intermolecular hydrogen bonds for each species. See Quantum Differences between Heavy and Light Water Physical Review Letters 101, 065502 for recent (2008) experimental data.
Numerous references characterized $\ce{D2O}$ as "more structured" than $\ce{H2O}$, meaning more hydrogen bonds, and a more narrow distribution of hydrogen bond lengths and angles. According to Effect of Ions on the Structure of Water: Structure Making and Breaking Chem. Rev. 2009, 109, 1346–1370 "It is indeed generally agreed that heavy water, $\ce{D2O}$, is more strongly hydrogen bonded (structured) than light water, $\ce{H2O}$." My explanation would therefore be that there is a greater penalty for placing ions in $\ce{D2O}$ than $\ce{H2O}$ as far as disruption of a hydrogen bonding network.
Explanation of Normal/Inverse Solvent Isotope Effect
For a kinetic normal/inverse solvent isotope effectthere will be a reactant and transition state. If (for example) there is a single solvent exchangeable proton that is the same group in the reactant and transition state, for example, $\ce{ROH}$ in the reactant and $\ce{R'OH}$ in the transition state, switching solvents from $\ce{H2O}$ to $\ce{D2O}$ will either favor the reactant or the transition state relative to each other (considering the respective $\ce{OH}$ bond strengths as well and intermolecular hydrogen bonds to solvent). IF $\ce{D2O}$ favors the reactant relative to the transition state (activation energy is increased), this is a "normal kinetic solvent isotope effect". Oppositely, if $\ce{D2O}$ favors the transition state relative to the reactant this is an "inverse kinetic solvent isotope effect." More complex scenarios involving more exchangeable sites can of course occur.
Similarly there can be equilibrium normal/inverse solvent isotope effect, if there is an equilibrium reaction and then it is reactant vs. product (rather than reactant vs. transition state) that matters.
Well, if you are interested in organic chemistry, you might be aware of the fact that the hyperconjugation effect of -CH3 is greater than -CD3. The reason for this is that the atoms in the molecules keep vibrating and the bonds of H being slightly weaker than D are easily delocalised. (due to heavier weight of D than H, the vibrations are weaker and thus the bonds are stable). The same is the case with water molecules where the bonds between O and H are weaker than O and D bond. Hence H2O is slightly acidic than D2O.
Well, if you are interested in organic chemistry, you might be aware of the fact that the hyperconjugation effect of -CH3 is greater than -CD3. The reason for this is that the atoms in the molecules keep vibrating and the bonds of H being slightly weaker than D are easily delocalised. (due to heavier weight of D than H, the vibrations are weaker and thus the bonds are stable). The same is the case with water molecules where the bonds between O and H are weaker than O and D bond. Hence H2O is slightly acidic than D2O.
This early paper reports that acidic compounds (phenols, carboxylic acids, and others) are noticeably more dissociated in $\ce{H2O}$ than $\ce{D2O}$. Since water (protio or deuterio) is just another example of a weak acid, it is not unreasonable to expect $\ce{H2O}$ and $\ce{D2O}$ to follow this pattern. This would suggest that the equilibrium constant in eqn I would be larger than the equilibrium constant in eqn II. That is, $\ce{H2O}$ in $\ce{H2O}$ should be more ionized than $\ce{D2O}$ in $\ce{D2O}$
The reason why this is the case is the interesting question.
Since equilibria are governed by thermodynamic (rather than kinetic) effects, we must examine the relative stabilities of the reactants and products in these two equations in order to determine the relative positions of these two equilibria.
On the reactant side, $\ce{D2O}$ is more stable than $\ce{H2O}$ since an $\ce{O-D}$ bond will have a lower zero-point energy than the corresponding $\ce{O-H}$ bond. Hence the deuterated molecule will have a lower overall energy compared to its protonated analogue.
On the product side, the deuterated ion must again be more stable than the protonated ion for the same zero-point energy reasons. However, since the bonds in the $\ce{X3O^+}$ ions are weaker (longer) than the bonds in the neutral reactants, the zero-point energy differences would be compressed or lessened.
The reactant effects discussed in the first bullet will tend to push equilibrium II further to the left than equilibrium I. The product effects discussed in the second bullet will tend to push equilibrium II further to the right than equilibrium I. In other words, the two effects are in opposition, but as explained in the second bullet, the product effect is likely smaller than the reactant effect so that overall the reactant effects win out and equilibrium II lies further to the left than equilibrium I.
This early paper reports that acidic compounds (phenols, carboxylic acids, and others) are noticeably more dissociated in $\ce{H2O}$ than $\ce{D2O}$. Since water (protio or deuterio) is just another example of a weak acid, it is not unreasonable to expect $\ce{H2O}$ and $\ce{D2O}$ to follow this pattern. This would suggest that the equilibrium constant in eqn I would be larger than the equilibrium constant in eqn II. That is, $\ce{H2O}$ in $\ce{H2O}$ should be more ionized than $\ce{D2O}$ in $\ce{D2O}$
The reason why this is the case is the interesting question.
Since equilibria are governed by thermodynamic (rather than kinetic) effects, we must examine the relative stabilities of the reactants and products in these two equations in order to determine the relative positions of these two equilibria.
On the reactant side, $\ce{D2O}$ is more stable than $\ce{H2O}$ since an $\ce{O-D}$ bond will have a lower zero-point energy than the corresponding $\ce{O-H}$ bond. Hence the deuterated molecule will have a lower overall energy compared to its protonated analogue.
On the product side, the deuterated ion must again be more stable than the protonated ion for the same zero-point energy reasons. However, since the bonds in the $\ce{X3O^+}$ ions are weaker (longer) than the bonds in the neutral reactants, the zero-point energy differences would be compressed or lessened.
The reactant effects discussed in the first bullet will tend to push equilibrium II further to the left than equilibrium I. The product effects discussed in the second bullet will tend to push equilibrium II further to the right than equilibrium I. In other words, the two effects are in opposition, but as explained in the second bullet, the product effect is likely smaller than the reactant effect so that overall the reactant effects win out and equilibrium II lies further to the left than equilibrium I.
Ya. Thats what $\mu$ represents. Its a way of reducing a two body problem into a one body problem which is commonly used for vibrating systems. Essentially youre looking at the vibration of the center of mass rather than the vibration of the two atoms. So, all bonds have a reduced mass not just OD. Sorry if that was unclear.More
For the reasons explained in New point of view on the meaning and on the values of $K_\mathrm{a}(\ce{H3O+, H2O})$ and $K_\mathrm{b}(\ce{H2O, OH-})$ pairs in water Analyst, February 1998, Vol. 123 (409–410), the $\mathrm{p}K_\mathrm{a}$ of $\ce{H3O+}$ in $\ce{H2O}$ and the $\mathrm{p}K_\mathrm{a}$ of $\ce{D3O+, D2O}$ are undefined.
The entire point of the above reference is that
$\ce{H3O+ + H2O <=> H2O + H3O+}$
(which would correspond to an equilibrium constant of 1) is not a genuine thermodynamic process because the products and reactants are the same.
$\ce{D3O+ + D2O <=> D2O + D3O+}$
would also correspond to an equilibrium constant of 1
So when Clayden and the OP write
it is wrong for the above reason.
Two genuine thermodynamic equilibriums are
$\ce{2H2O <=> H3O+ + HO-}$ and $\ce{2D2O <=> D3O+ + DO-}$
Experimentally, the self-dissociation constants of $\ce{H2O}$ to $\ce{H3O+}$ and $\ce{OH-}$ and $\ce{D2O}$ to $\ce{D3O+}$ and $\ce{OD-}$ can be measured as in The Ionization Constant of Deuterium Oxide from 5 to 50 [degrees] J. Phys. Chem., 1966, 70, pp 3820–3824 and it is found that $\ce{H2O}$ is about 8 times more dissociated (equilibrium constant is 8 times greater).
But using the above data to say $\ce{D3O+}$ is stronger is misleading, because this corresponds to a reaction with $\ce{OD-}$, not $\ce{D2O}$. $\ce{D3O+}$ simply has a lower concentration in heavy water than $\ce{H3O+}$ has in light water.
As for why the $\ce{D2O}$ is less dissociated than $\ce{H2O}$, The ionization constant of heavy water ($\ce{D2O}$) in the temperature range 298 to 523 K Canadian Journal of Chemistry, 1976, 54(22): 3553-3558 breaks the differences down in to enthalpy and entropy components, which both favor ionization of $\ce{H2O}$ and states that $\ce{D2O}$ is a more structured liquid than $\ce{H2O}$. Not only the bonds of each product and reactant molecule need to be considered, but also the intermolecular forces: the number and strength of intermolecular hydrogen bonds for each species. See Quantum Differences between Heavy and Light Water Physical Review Letters 101, 065502 for recent (2008) experimental data.
Numerous references characterized $\ce{D2O}$ as "more structured" than $\ce{H2O}$, meaning more hydrogen bonds, and a more narrow distribution of hydrogen bond lengths and angles. According to Effect of Ions on the Structure of Water: Structure Making and Breaking Chem. Rev. 2009, 109, 1346–1370 "It is indeed generally agreed that heavy water, $\ce{D2O}$, is more strongly hydrogen bonded (structured) than light water, $\ce{H2O}$." My explanation would therefore be that there is a greater penalty for placing ions in $\ce{D2O}$ than $\ce{H2O}$ as far as disruption of a hydrogen bonding network.
Also the equilibrium constant for
$\ce{H2O + H2DO+ <=> HDO + H3O+}$
can be measured and it is 0.96 according to Isotopic Fractionation of Hydrogen between Water and the Aqueous Hydrogen Ion J. Phys. Chem., 1964, 68 (4), pp 744–751
Explanation of Normal/Inverse Solvent Isotope Effect
For a kinetic normal/inverse solvent isotope effect there will be a reactant and transition state. If (for example) there is a single solvent exchangeable proton that is the same group in the reactant and transition state, for example, $\ce{ROH}$ in the reactant and $\ce{R'OH}$ in the transition state, switching solvents from $\ce{H2O}$ to $\ce{D2O}$ will either favor the reactant or the transition state relative to each other (considering the respective $\ce{OH}$ bond strengths as well and intermolecular hydrogen bonds to solvent). IF $\ce{D2O}$ favors the reactant relative to the transition state (activation energy is increased), this is a "normal kinetic solvent isotope effect". Oppositely, if $\ce{D2O}$ favors the transition state relative to the reactant this is an "inverse kinetic solvent isotope effect." More complex scenarios involving more exchangeable sites can of course occur.
Similarly there can be equilibrium normal/inverse solvent isotope effect, if there is an equilibrium reaction and then it is reactant vs. product (rather than reactant vs. transition state) that matters.
For the reasons explained in New point of view on the meaning and on the values of $K_\mathrm{a}(\ce{H3O+, H2O})$ and $K_\mathrm{b}(\ce{H2O, OH-})$ pairs in water Analyst, February 1998, Vol. 123 (409–410), the $\mathrm{p}K_\mathrm{a}$ of $\ce{H3O+}$ in $\ce{H2O}$ and the $\mathrm{p}K_\mathrm{a}$ of $\ce{D3O+, D2O}$ are undefined.
The entire point of the above reference is that
$\ce{H3O+ + H2O <=> H2O + H3O+}$
(which would correspond to an equilibrium constant of 1) is not a genuine thermodynamic process because the products and reactants are the same.
$\ce{D3O+ + D2O <=> D2O + D3O+}$
would also correspond to an equilibrium constant of 1
So when Clayden and the OP write
it is wrong for the above reason.
Two genuine thermodynamic equilibriums are
$\ce{2H2O <=> H3O+ + HO-}$ and $\ce{2D2O <=> D3O+ + DO-}$
Experimentally, the self-dissociation constants of $\ce{H2O}$ to $\ce{H3O+}$ and $\ce{OH-}$ and $\ce{D2O}$ to $\ce{D3O+}$ and $\ce{OD-}$ can be measured as in The Ionization Constant of Deuterium Oxide from 5 to 50 [degrees] J. Phys. Chem., 1966, 70, pp 3820–3824 and it is found that $\ce{H2O}$ is about 8 times more dissociated (equilibrium constant is 8 times greater).
But using the above data to say $\ce{D3O+}$ is stronger is misleading, because this corresponds to a reaction with $\ce{OD-}$, not $\ce{D2O}$. $\ce{D3O+}$ simply has a lower concentration in heavy water than $\ce{H3O+}$ has in light water.
As for why the $\ce{D2O}$ is less dissociated than $\ce{H2O}$, The ionization constant of heavy water ($\ce{D2O}$) in the temperature range 298 to 523 K Canadian Journal of Chemistry, 1976, 54(22): 3553-3558 breaks the differences down in to enthalpy and entropy components, which both favor ionization of $\ce{H2O}$ and states that $\ce{D2O}$ is a more structured liquid than $\ce{H2O}$. Not only the bonds of each product and reactant molecule need to be considered, but also the intermolecular forces: the number and strength of intermolecular hydrogen bonds for each species. See Quantum Differences between Heavy and Light Water Physical Review Letters 101, 065502 for recent (2008) experimental data.
Numerous references characterized $\ce{D2O}$ as "more structured" than $\ce{H2O}$, meaning more hydrogen bonds, and a more narrow distribution of hydrogen bond lengths and angles. According to Effect of Ions on the Structure of Water: Structure Making and Breaking Chem. Rev. 2009, 109, 1346–1370 "It is indeed generally agreed that heavy water, $\ce{D2O}$, is more strongly hydrogen bonded (structured) than light water, $\ce{H2O}$." My explanation would therefore be that there is a greater penalty for placing ions in $\ce{D2O}$ than $\ce{H2O}$ as far as disruption of a hydrogen bonding network.
Also the equilibrium constant for
$\ce{H2O + H2DO+ <=> HDO + H3O+}$
can be measured and it is 0.96 according to Isotopic Fractionation of Hydrogen between Water and the Aqueous Hydrogen Ion J. Phys. Chem., 1964, 68 (4), pp 744–751
Explanation of Normal/Inverse Solvent Isotope Effect
For a kinetic normal/inverse solvent isotope effectthere will be a reactant and transition state. If (for example) there is a single solvent exchangeable proton that is the same group in the reactant and transition state, for example, $\ce{ROH}$ in the reactant and $\ce{R'OH}$ in the transition state, switching solvents from $\ce{H2O}$ to $\ce{D2O}$ will either favor the reactant or the transition state relative to each other (considering the respective $\ce{OH}$ bond strengths as well and intermolecular hydrogen bonds to solvent). IF $\ce{D2O}$ favors the reactant relative to the transition state (activation energy is increased), this is a "normal kinetic solvent isotope effect". Oppositely, if $\ce{D2O}$ favors the transition state relative to the reactant this is an "inverse kinetic solvent isotope effect." More complex scenarios involving more exchangeable sites can of course occur.
Similarly there can be equilibrium normal/inverse solvent isotope effect, if there is an equilibrium reaction and then it is reactant vs. product (rather than reactant vs. transition state) that matters.
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Well, if you are interested in organic chemistry, you might be aware of the fact that the hyperconjugation effect of -CH3 is greater than -CD3. The reason for this is that the atoms in the molecules keep vibrating and the bonds of H being slightly weaker than D are easily delocalised. (due to heavier weight of D than H, the vibrations are weaker and thus the bonds are stable). The same is the case with water molecules where the bonds between O and H are weaker than O and D bond. Hence H2O is slightly acidic than D2O.
Well, if you are interested in organic chemistry, you might be aware of the fact that the hyperconjugation effect of -CH3 is greater than -CD3. The reason for this is that the atoms in the molecules keep vibrating and the bonds of H being slightly weaker than D are easily delocalised. (due to heavier weight of D than H, the vibrations are weaker and thus the bonds are stable). The same is the case with water molecules where the bonds between O and H are weaker than O and D bond. Hence H2O is slightly acidic than D2O.
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This early paper reports that acidic compounds (phenols, carboxylic acids, and others) are noticeably more dissociated in $\ce{H2O}$ than $\ce{D2O}$. Since water (protio or deuterio) is just another example of a weak acid, it is not unreasonable to expect $\ce{H2O}$ and $\ce{D2O}$ to follow this pattern. This would suggest that the equilibrium constant in eqn I would be larger than the equilibrium constant in eqn II. That is, $\ce{H2O}$ in $\ce{H2O}$ should be more ionized than $\ce{D2O}$ in $\ce{D2O}$
\begin{aligned} \ce{H2O + H2O &~<=> H3O^+ + OH^{-}&&(I)}\\ \ce{D2O + D2O &~<=> D3O^+ + OD^{-}&&(II)}\\ \end{aligned}
The reason why this is the case is the interesting question.
Since equilibria are governed by thermodynamic (rather than kinetic) effects, we must examine the relative stabilities of the reactants and products in these two equations in order to determine the relative positions of these two equilibria.
The reactant effects discussed in the first bullet will tend to push equilibrium II further to the left than equilibrium I. The product effects discussed in the second bullet will tend to push equilibrium II further to the right than equilibrium I. In other words, the two effects are in opposition, but as explained in the second bullet, the product effect is likely smaller than the reactant effect so that overall the reactant effects win out and equilibrium II lies further to the left than equilibrium I.
This early paper reports that acidic compounds (phenols, carboxylic acids, and others) are noticeably more dissociated in $\ce{H2O}$ than $\ce{D2O}$. Since water (protio or deuterio) is just another example of a weak acid, it is not unreasonable to expect $\ce{H2O}$ and $\ce{D2O}$ to follow this pattern. This would suggest that the equilibrium constant in eqn I would be larger than the equilibrium constant in eqn II. That is, $\ce{H2O}$ in $\ce{H2O}$ should be more ionized than $\ce{D2O}$ in $\ce{D2O}$
\begin{aligned}\ce{H2O + H2O &~<=> H3O^+ + OH^{-}&&(I)}\\\ce{D2O + D2O &~<=> D3O^+ + OD^{-}&&(II)}\\\end{aligned}
The reason why this is the case is the interesting question.
Since equilibria are governed by thermodynamic (rather than kinetic) effects, we must examine the relative stabilities of the reactants and products in these two equations in order to determine the relative positions of these two equilibria.
The reactant effects discussed in the first bullet will tend to push equilibrium II further to the left than equilibrium I. The product effects discussed in the second bullet will tend to push equilibrium II further to the right than equilibrium I. In other words, the two effects are in opposition, but as explained in the second bullet, the product effect is likely smaller than the reactant effect so that overall the reactant effects win out and equilibrium II lies further to the left than equilibrium I.
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