The base dissociation equilibrium constant (Kb) for C5H5N is 1.40×10-9 (eq. 1).
(eq. 1): C5H5N(aq) + H2O(l) = OH-(aq) + C5H5NH+(aq)
Calculate the value of the equilibrium constant (Ka) for the hydrolysis of C5H5NH+ as shown in the reaction (eq. 2).
(eq. 2): C5H5NH+(aq) + H2O(l) = H3O+(aq) + C5H5N(aq)
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To find the equilibrium constant (Ka) for the hydrolysis of C5H5NH+, we can use the relationship between Kb and Ka for a conjugate acid-base pair.
The Kb expression for the reaction in eq. 1 is:
Kb = [OH-][C5H5NH+]/[C5H5N][H2O]
Since water concentration is constant, we can write Kb as:
Kb = [OH-][C5H5NH+]/[C5H5N]
Now, let's look at the hydrolysis reaction in eq. 2:
C5H5NH+(aq) + H2O(l) = H3O+(aq) + C5H5N(aq)
By comparing eq. 2 to eq. 1, we can see that the hydrolysis reaction in eq. 2 is the reverse of the reaction in eq. 1.
Therefore, we can write the equilibrium constant expression (Ka) for eq. 2 as the reciprocal of Kb:
Ka = 1/Kb
Let's calculate Ka using this relationship:
Ka = 1/(1.40×10-9)
Ka ≈ 7.14 × 108
No problem! I'm here to help. To calculate the value of the equilibrium constant (Ka) for the reaction in equation 2, we can use the relationship between Ka and Kb for a conjugate acid-base pair.
The Ka and Kb values for a conjugate acid-base pair are related by the equation:
Kw = Ka * Kb
Where Kw is the equilibrium constant for the autoionization of water, which is equal to 1.0 x 10^-14 at 25°C.
In equation 1, the base dissociation equilibrium constant (Kb) for C5H5N is given as 1.40 x 10^-9. Using this value, we can calculate the corresponding value of Ka for the hydrolysis reaction in equation 2.
First, we need to determine the value of Kb for the conjugate base OH- in equation 1. Since Kw = Ka * Kb, we can rearrange the equation to solve for Kb:
Kb = Kw / Ka
Substituting the known values, we get:
Kb = (1.0 x 10^-14) / (1.40 x 10^-9)
Now that we have the value of Kb for OH-, we can use it to calculate Ka for the reaction in equation 2. Since Ka and Kb are related as conjugate acid-base pairs, we can simply switch the values:
Ka = Kw / Kb
Substituting the value of Kb, we get:
Ka = (1.0 x 10^-14) / Kb
Calculating this expression will give us the value of Ka for the hydrolysis reaction in equation 2.
Ka = (Kw/Kb).
Just remember that KaKb = Kw. The Ka and Kb are interchangeable with that formula.