what is the hydronium-ion concentration at equilibrium in a 0.71 M solution of aniline (Kb=4.2x10^-10) at 25 Celsius
See your post on CH3NH2.
To determine the hydronium-ion concentration at equilibrium in a solution of aniline, we need to use the given Kb value.
First, let's write the balanced chemical equation for the dissociation of aniline:
C6H5NH2 + H2O ⇌ C6H5NH3+ + OH-
The Kb expression for this reaction is:
Kb = ([C6H5NH3+][OH-]) / [C6H5NH2]
Since the Kb value is given as 4.2 x 10^-10, we can plug it into the expression:
4.2 x 10^-10 = ([C6H5NH3+][OH-]) / [C6H5NH2]
In this problem, we are given the concentration of aniline, which is 0.71 M. However, we're interested in finding the equilibrium concentration of the hydronium ion (H3O+), so we need to solve for that.
To do this, we can use the fact that for any aqueous solution at 25°C, [H3O+] x [OH-] = 1.0 x 10^-14 (the ion product of water). Therefore, [OH-] = 1.0 x 10^-14 / [H3O+].
Substituting this into the Kb expression, we have:
4.2 x 10^-10 = ([C6H5NH3+][(1.0 x 10^-14) / [H3O+]]) / [C6H5NH2]
Now we can plug in the values:
4.2 x 10^-10 = ([H3O+][(1.0 x 10^-14) / [H3O+]]) / 0.71
Simplifying, we get:
4.2 x 10^-10 = (1.0 x 10^-14) / 0.71
Now, rearrange the equation to solve for [H3O+]:
[H3O+] = (4.2 x 10^-10) x 0.71 / (1.0 x 10^-14)
Calculating this expression will give you the hydronium-ion concentration at equilibrium in the 0.71 M solution of aniline.