Question

This question has multiple parts. Work all the parts to get the most points. A titration of 21.7 mL of a solution of the weak base aniline, CH NH, (K) = 4.0 x 10-1), requires 24.35 mL of 0.175 M HCl to reach the equivalence point. C.H NH, (aq) + H, 0+ (aq) = CH NH,+ (aq) + H, 0(1) a What was the concentration of aniline in the original solution? Concentration - 196 MV The equivalence point for a reaction is the point at which one reactant has been completely consumed by addition of another reactant. Thus, 0.175 mol HCL x 24.35 mL X - 1 mol C6H NHA ICH NH,] -0.175 mol HCI 1.00 L 1 mol HCI * 21.7 ml. -0.196 M What are the concentrations of H, 0+, OH, and C. H NH3 at the equivalence point? 1,0+) - (OH)- [C,H, NH*]= Submit Submit Answer Submit Answer Try Another Version 6 Item attempts remaining
just b please thank you.

Answer 1

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

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

Now, number of moles of HCl added is

0.175 mol " x 0.02435 L 0.00426 mol 1 L

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.

C6H5NH,

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

Hence, the concentration of formed is

C6H5NH,

0.00426 mol C6H5NH] = - 0.04605 L 0.0925 M

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

C6H5NH,

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

Given that Kb of aniline is

K; = 4.0 x 10-10

We can calculate the Ka of as follows

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

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

	C6H5NH]	C6H5NH2	40*НІ
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:

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.

Hence, the equilibrium concentrations are

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

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

[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

Hence, the equilibrium concentrations at equivalence point are

[H3O+] = 1.51 x 10-3 M

OH-] = 6.63 x 10-12 M

[C6H5NH2) = 9.10 x 10-2 M