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 

Answer 2

The pH of the solutions after adding 0.02mol of HCl will be 1.7, 1.69, 1.7, and 6.4 respectively.

pH is used to measure the alkalinity and acidity of the solutions. pH 7 is neutral.

For solution A,

$\bold {Kb =\dfrac {[OH^-][HONH3+]}{[NH2OH]} }$

put the value and solve it for [OH]

$\bold {[OH^-] = 6.6x10^-^5 M }$

$\bold{ pOH = -log[OH]}\\\\\bold{ pOH = -log(6.6x10^-^5)}\\\\\bold {pOH = 4.18}}$

Since, pH = 14 - pOH

pH = 9.82

Since, $\bold{[H^+] = 0.02}$

$\bold {pH = -log (0.02)}\\\bold {pH = 1.7}$

For Solution B

$\bold {Ka [HONH_3Cl ]= 9.09x10^-7 }\\\\\bold {Ka = \dfrac { [H+][HONH2]} { [HONH3+]}}$

Since, $\bold { [H^+] = 6x10^-^4 }$

So, pH  =$\bold {pH = -log (6x10^-4) = 3.22}$

The new pH will be

$\bold {pH = -log (6x10^-4+0.02)}\\\bold {pH = 1.69}$

(C) In pure wate, HCl completely dissociates, thus the pH will be

$\bold{[H^+] = 0.02}\\\bold {pH = -log (0.02)}\\\bold {pH = 1.7}$

(D) Since, $\bold {HONH_2}$ and $\bold {HONH_3Cl}$ are at same concentration.

So, $\bold {pH = pKa}$

$\bold {Pka = -log Ka}\\\\\bold {pKa = -log 9.09x10^-7}\\\\\bold{ pKa = 6.04}\\\\\bold{ pH = 6.04}$

Therefore, The pH of the solutions after adding 0.02mol of HCl will be 1.7, 1.69, 1.7, and 6.4 respectively.

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