Question 18 Not yet answered Marked out of 1.00 P Flag question Find the pH of 0
ID: 1090552 • Letter: Q
Question
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 questionExplanation / Answer
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
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