OH- + C6H5CH2COOH H2O + C6H5CH2COO-
What is the K for the reaction above in terms of Ka's, Kb's, Kw etc.?
The arrow goes after the C6H5CH2COOH
Answered below AND I corrected the typo there, too.
To determine the equilibrium constant (K) for the given reaction, you need to consider the acid dissociation constant (Ka) and the base dissociation constant (Kb) for the species involved. In this case, we are given the equation for the reaction between a weak acid (C6H5CH2COOH) and a strong base (OH-).
First, let's consider the dissociation of the weak acid C6H5CH2COOH:
C6H5CH2COOH ⇌ C6H5CH2COO- + H+
The corresponding expression for the acid dissociation constant (Ka) is as follows:
Ka = [C6H5CH2COO-][H+]/[C6H5CH2COOH]
Next, let's consider the dissociation of water (H2O):
H2O ⇌ H+ + OH-
The dissociation constant for water (Kw) is defined as:
Kw = [H+][OH-]/[H2O]
Since the reaction involves the strong base OH-, we can consider OH- as the product of water's dissociation:
OH- = Kw/[H+]
Now, substitute [OH-] and [H+] in terms of Kw into the acid dissociation expression:
Ka = ([C6H5CH2COO-]*Kw)/([H+]^2)
Finally, we can rewrite the equation for the given reaction using the expression for Ka:
OH- + C6H5CH2COOH ⇌ C6H5CH2COO- + H2O
Dividing the equation by [C6H5CH2COOH] and rearranging the terms, we obtain:
[C6H5CH2COO-]*([H+]^2)/(Kw) = [OH-]
Now, we can substitute this into the original equation for Ka:
Ka = ([OH-]*Kw)/[C6H5CH2COOH]
So, the equilibrium constant (K) for the given reaction is Ka = ([OH-]*Kw)/[C6H5CH2COOH].