Question

Determine the pH of each of the following solutions. I've already tried taking the -log, and subtracting the -log from 14 for bases, but it still tells me im getting it wrong. These are answers i got that are aparently incorrect. Can anyone help?

0.255 M CH3NH3I pH = 13.4

0.316 M KC6H5O pH = 13.5

0.160 M HNO2 pH = 0.8

Answer 1

0.255 M CH3NH3I pH = 13.4

CH3NH3I ---> CH3NH3+ + I-

CH3NH3+ + H2O <--> CH3NH2 + H3O+

Ka = [CH3NH2] [H3O+] / [CH3NH3+]

expect acidic pH; you have basic, thats why you are not correct with pH = 13.4

pKb = 3.36

pKa = 14-3.36 = 10.64

Ka = 10^-10.64

Apply

Ka = [CH3NH2] [H3O+] / [CH3NH3+]

Ka = x*x/(M-x)

10^-10.64 = x*x/(0.255 -x)

since ka is too small then

10^-10.64 = x*x/(0.255)

x = sqrt((10^-10.64)*(0.255)) = 0.00000241696

pH = -log(0.00000241696) =5.61673053698

using smae logic

for phenol

pKa = 10

pKb = 4

x = [OH-] = sqrt(Kb*M) = sqrt((10^-4)*0.316 ) = 0.00562138772

pOH = -log(0.00562138772) =2.2501

ph = 14-2.2501

pH = 11.7499

HNO2 is a weak acid

Ka = [h+][NO2-]/[HNO2]

Ka= 4*10^-4

assume

[h+]= x = [NO2-]

[HNO2] = 0.16 -x

Ka = [h+][NO2-]/[HNO2]

4*10^-4= x*x/(0.16 -x)

x = 0.007802499609497022

[H+] = 0.007802499609497022

pH = -log(0.007802499609497022 =2.1