Just b please thank you. This question has multiple parts. Work all the parts to get...

We have 21.7 mL of aniline to which 24.35 mL(=0.02435 L) of 0.175 M HCl was added to reach the equivalence point.

The reaction that happens is

Now, number of moles of HCl added is

Since this addition resulted in reaching the equivalence point, the starting moles of aniline must be 0.00426 mol.

Hence, in total we will have 0.00426 mol of after reaction.

Total volume of the solution = 21.7 mL + 24.35 mL = 46.05 mL = 0.04605 L

Hence, the concentration of formed is

Note that is an weak acid that will dissociate as follows:

Given that Kb of aniline is

We can calculate the Ka of as follows

Now, we can create the following ICE chart to calculate the equilibrium number of moles of each species present according to the above reaction.

Initial, M	0.0925	0	0
Change, M	-x	+x	+x
Equilibrium, M	0.0925-x	x	x

Hence, we can write the Ka expression for dissociation from the above ICE chart as follows:

Hence, the equilibrium concentrations are

Now, from H₃O⁺ concentration, we can find the OH- concentration as follows:

Hence, the equilibrium concentrations at equivalence point are

C6H5NH2(aq) + H2O (from HCI) + CHEN H(aq) + H200

0.175 mol " x 0.02435 L 0.00426 mol 1 L

C6H5NH,

C6H5NH,

0.00426 mol C6H5NH] = - 0.04605 L 0.0925 M

C6H5NH,

C6H5NH3(aq) + H200 + C6H5NH2(aq) + H30

K; = 4.0 x 10-10

C6H5NH,

KaxKb = 1.0 x 10-14 1.0 x 10-14 ha= 4.0 X 10-1 4.0 x 10-10 = 2.5 x 10-5

C6H5NH]

C6H5NH2

40*НІ

C6H5NH,

Ka= H30+][C6H5NH2] XXX CH5NH,110,0925 --- =2.5 x 10-5 = 2 = 2.5 x 10-5 (0.0925 – ) 1.51 x 10-3 M.

H30+1 = 1 = 1.51 x 10-3 M = 0.00151 M

C6H5NH] = 0.0925 M - r = 0.0925 M – 1.51 x 10-3 M C6H5NH10.0910 M = 9.10 x 10-2 M

[OH-][H30+1 = 1.00 x 10-14 1.00 x 10-14 → [OH-]= 1.51 x 10-3 = 6.63 x 10-12 M

[H3O+] = 1.51 x 10-3 M

OH-] = 6.63 x 10-12 M

[C6H5NH2) = 9.10 x 10-2 M