Calculate the concentration of all species in a 0240 M C6H5N

Calculate the concentration of all species in a 0.240 M C6H5NH3Cl solution.

[C6H5NH3+], [Cl-], [C6H5NH2], [H3O+], [OH-]

Solution

C6H5NH3Cl is a strong salt. so it undergoes complete dissociation

C6H5NH3Cl= 0.240M

C6H5NH3Cl ---------------- C6H5NH3+ + Cl-

[Cl-]= 0.240M

C6H5NH3+ H2O ------------------ C6H5NH2 + H3O+

0.240                                            0               0

-x                                              +x             +x

0.240-x                                       +x             +x

Ka= [C6H5NH2][H3O+]/[C6H5NH3+]               Ka= 2.6x10^-5

2.6x10^-5 = x*x/(0.240-x)

for solving x= 0.0025

[C6H5NH2] = x= 0.0025M

[H3O+]= 0.0025M

[C6H5NH3+] = 0.240-x= 0.240 - 0.0025= 0.2375= 0.236M

[H3O+] = 0.0025M

[H3O+]{OH-]= Kw               Kw= ionic product of water= 1.0x10^-14

[OH-] = Kw/[H3O+]

[OH-] = 1.0x10^-14/0.0025= 4.0x10^-12M

[OH-] = 4.0x10^-12M