I have an assignment Q
A 0.1 mol L-1 water solution of NH4CH3CO2 has a pH of approx 7. Explain this using two equations.
I have one equation
CH3COO-(aq) + H2O(l) --> CH3COOH(aq) + OH-(aq)
What could the other equation be?
NH4CH3CO2 is the salt of a weak acid and a weak base.
The equation is OK for a weak acid, you need a similar one for a weak base.
To explain why the pH of the water solution of NH4CH3CO2 is approximately 7, we need to consider the dissociation of NH4CH3CO2 into its ions and the subsequent chemical reactions that occur in water.
First, let's write the equation for the dissociation of NH4CH3CO2:
NH4CH3CO2(aq) ⟶ NH4+(aq) + CH3COO-(aq)
Now, we have NH4+ ions, a weak acid, and CH3COO- ions, a weak base. To determine the other equation, we need to consider the hydrolysis of NH4+.
The hydrolysis of NH4+ can be represented by the following equation:
NH4+(aq) + H2O(l) ⟶ NH3(aq) + H3O+(aq)
In this equation, NH4+ reacts with water (H2O) to form NH3 (ammonia) and H3O+ (hydronium ion). Ammonia is a weak base while hydronium ion acts as an acid.
When both the weak acid (CH3COO-) and the weak base (NH4+) are present in a solution, both reactions occur simultaneously. These reactions help to neutralize each other, resulting in a solution that is close to neutral pH.
So, the second equation, representing the hydrolysis of NH4+, is:
NH4+(aq) + H2O(l) ⟶ NH3(aq) + H3O+(aq)
Together with the first equation you provided:
CH3COO-(aq) + H2O(l) ⟶ CH3COOH(aq) + OH-(aq)
These equations explain why the pH of a 0.1 mol L-1 water solution of NH4CH3CO2 is approximately 7. The weak acid (CH3COO-) reacts with water to produce hydronium ions (H3O+) and the weak base (NH4+) hydrolyzes to produce ammonia (NH3) and additional hydronium ions (H3O+), resulting in a near-neutral solution.