Question

Determine the pH of each of the following solutions.

A) 0.17 M CH3NH3I (Kb for CH3NH2 is 4.4×10−4): Express your answer to two decimal places.

B) 0.20 M KI: Express your answer to two decimal places.

Answer 1

Answer - We are given, molarity of weak acid [CH₃NH₃I] = [CH₃NH₃⁺] = 0.17 M

Kb = 4.4*10^-4

we need to put ICE table for calculating the [H₃O⁺]

    CH₃NH₃⁺+ H₂O ------>   H₃O⁺ +   CH₃NH₂

I   0.17                                 0                0

C -x                                    +x               +x

E 0.17-x                             +x               +x

We need to calculate Ka from Kb

Kb = Kw / Ka

       = 1 * 10^-14 / 4.4 * 10^-4

       = 2.27 *10^-11

Ka = [H₃O⁺] [CH₃NH₂] / [CH₃NH₃⁺]

2.27 *10^-11 = x*x /(0.17-x)

We know the Ka value is too small, so we can neglet x in the 0.17-x

32.27 *10^-11 * 0.17 = x²

x = 1.96 x10^-6 M

so, x = [H₃O⁺] = 1.96 x10^-6 M

so, pH = -log [H₃O⁺]

           = -log 1.96 x10^-6 M

           = 5.70

So the pH of 0.17 M CH₃NH₃I solution is 5.70

b) 0.20 M KI

[KI] = [I^-] = 0.20 M

We know KI is neutral salt since it is formed from strong acid HI and strong base KOH, so its pH is neutral

so, pH for 0.20 M KI solution is 7.00