Question

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Determine the pH of each of the following solutions. 0.21 M CH3NH3I pH = 11.97

Answer 1

pH of 0.21 M CH3NH3I

Solution:

We know the CH3NH3I is a salt of weak base and strong acid and its completely ionizes

CH3NH3I --> CH3NH3+ + I-

CH3NH3+ is acid weak acid and its ka value is calculated by using kb value of CH3NH3

Kb of CH3NH3 =4.2 X 10^-4

Ka = 1.0 x 10^-14 / kb

= 1.0 x 10^-14 / 4.2 X 10^-4

=2.381 x 10^-11

Ionization of weak acid and ICE chart

             CH3NH3⁺(aq)       + H2O (l)    --> CH3NH2 (aq) + H3O⁺(aq)

I             0.21                                                   0                      0

C          -x                                                         +x                    +x

E     (0.21-x)                                                    x                      x

Ka expression

Ka = [CH3NH2 (aq)][ H3O⁺(aq)]/ [CH3NH3⁺(aq) ]

Lets plug equilibrium concentration of each species

Ka = x²/ (0.21-x)

2.381 X 10^-11 = x²/ (0.21-x)

Ka value is too small and its less than 10^-5 . So we use 5% approximation

2.381 X 10^-11 = x²/ 0.21

X² = 2.381 X 10^-11 * 0.21

x = 2.24 x 10^-6

We know [H3O+] =x = 2.24 x 10^-6

We know pH = - log [H3O+]

= -log (2.24 x 10^-6 )

= 5.65

pH = 5.65