Question 18 Not yet answered Marked out of 100 P Flag questi

Question 18 Not yet answered Marked out of 1.00 P Flag question Find the pH of 0 135 M phenol, CoHsOH. The H bonded to the O is acidic, that is CHsOH(ag) H200) H3O (aq)+CsHso (aq) Ka 1.0 x 10-10 M Use the same procedure as used in the last problem Answer Question 19 Not yet answered Marked out of 1.00 ? Flag question Calculate the approximate [OH 1 and INHA 1in a 057 M ammonia solution, NHs(aq) NHs(aq)+ H2O) OH (aq) + NH4 (aq) Kb 1.75 x105M Use the same procedure as used in the last 2 problems Details for those who want a little extra help Calculate pH of weak base Answer Question 20 Calculate the pH of 0 194 M ammonia Not yet NHslag) + H2O0) OH (aq)+ NH4 (aq) K 175 x 105 Marked out of 1 00 Answer P Fiag question

Solution

Q.No.18.

C6H5OH + H2O ------------------- H3O+ + C6H5O-

0.135M                                   0             0

-x                                          +x           +x

0.135 -x                               +x                +x

Ka = [H3O+][C6H5O-]/[C6H5OH]

1.0x10^-10 = x*x/0.135-x

0.135x10^-10 = x^2

x= 0.367 x10^-5

[H3O+] = 0.367x10^-5M

-log[H3O+] = -log[0.367x10^-5]

PH= 5.43.

QNo.19

NH3 + H2O ----------- OH- + NH4+

0.57                       0              0

-x                            +x          +x

0.57-x                   +x            +x

kb= [OH-][NH4+]/[NH3]

1.75x10^-5 = x*x/0.57-x

for solving this equation

x=0.0031

[OH-] = x= 0.0031M= 3.1x10^-3M

[NH4+] = x= 0.0031M= 3.1x10^-3M

QNo.20

NH3 + H2O -------------- OH-   + NH4+

0.194                          0          0

-x                             +x          +x

0.194-x                       +x         +x

kb= [OH-][MH4+]/[NH3]

1.75x10^-5 = x*x/0.194-x

for solving this equation

x= 0.0018

[OH-] = 0.0018M

-log[OH-] -log[0.0018]

POH= 2.74

PH+POH= 14

PH= 14-POH

PH= 14-2.74

PH= 11.26