We’re being asked to calculate the pH of a 0.16 M CH3NH3Cl using an ICE chart.
CH3NH3+ is a conjugate acid of the base CH3NH2. We need to use the Kb for CH3NH2 to find the Ka of its conjugate acid, CH3NH3+. Let us just omit Cl- since it is neutral.
We can find on books and the internet that the Kb of CH3NH2 is 4.4x10-4. Therefore, the Ka of CH3NH3+ is:
Ka = 2.27x10-11
Since CH3NH3+ has a very low Ka value, it’s a weak acid. Remember that weak acids partially dissociate in water and that acids donate H+ to the base (water in this case). The dissociation of CH3NH3+ is as follows:
CH3NH3+(aq) + H2O(l) ⇌ H3O+(aq) + CH3NH2(aq); Ka = 2.27×10–11
From this, we can construct an ICE table. Remember that liquids are ignored in the ICE table.
The Ka expression for CH3NH3+ is:
Liquids are ignored in the Ka expression.
Note that each concentration is raised by the stoichiometric coefficient: [CH3NH3+], [H3O+] and [CH3NH2] are raised to 1.
Solve an equilibrium problem (using an ICE table) to calculate the pH of each of the following solutions.
0.16 M CH3 NH3 Cl
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Our tutors have indicated that to solve this problem you will need to apply the Weak Acids concept. If you need more Weak Acids practice, you can also practice Weak Acids practice problems.