Question

Calculate the ph after 0.020 mol hcl is added to 1.00 l of each of the four solutions below.

Answer 1

A) according to this reaction:
by using ICE table:
              NH2OH(aq) + H2O(l) → HONH3+(aq)   + OH-
initial       0.4 M                                       0                    0
change     -X                                          +X                  +X
Equ        (0.4-X)                                         X                    X

when Kb = [OH-][HONH3+]/[NH2OH]  
when we have Kb = 1.1x10^-8 so,
by substitution:
1.1x10^-8 = X^2/(0.4-X) by solving this equation for X 

∴X = 6.6x10^-5 M
∴[OH] = 6.6x10^-5 M 
when POH = - ㏒[OH]
    ∴POH = -㏒(6.6x10^-5)= 4.18
∴PH = 14 - POH = 14 - 4.18
        = 9.82
when PH = -㏒[H+]
∴[H+] = 10^9.82 = 1.5x10^-10 M+0.02molHcl
          = 0.02
∴ the new value of PH = -㏒(0.02)
∴PH = 1.7

B) according to this reaction:
 by using ICE table:
             HONH3+(aq) → H+(aq) + HONH2(aq)
intial     0.4                          0            0
change -X                          +X           +X
Equ       (0.4-X)                    X              X

when Ka HONH3Cl = 9.09x10^-7 
and Ka = [H+][HONH2] / [HONH3+]

So by substitution and we can assume [HONH3+] = 0.4 as the value of Ka is so small so,
9.09x10^-7 = X^2 / 0.4 by solving for X
∴  X = 6 x 10 ^-4
∴[H+] = 6x10^-4
PH = -㏒[H+] 
      = -㏒ (6x10^-4) = 3.22
when [H+] = 6x10^-4 + 0.02 m HCl 
∴new value of PH = -㏒(6x10^-4+0.02)
                               = 1.69
C) when we have pure H2O and PH of water = 7
So we can get [H+] when PH = -㏒[H+]
∴[H+] = 10^-7 + 0.02MHCl
          = 0.02
∴new value of PH = -㏒0.02
                         PH = 1.7
d) when HONH2 & HONH3Cl have the same concentration and Hcl added to them so we can assume that PH=Pka
and when we have Ka for HONH3Cl = 9.09x10^-7 
So we can get the Pka:

Pka = -㏒Ka
       = -㏒9.09x10^-7
       = 6.04
∴PH = 6.04
and because of the concentration of the buffer components, HONH2 & HONH3Cl have 0.4 M and the adding of HCl = 0.02 M So PH will remain very near to 6