calculate pH of 0.1M NH4^- of 100ml of that solution?
To calculate the pH of a solution, we need to consider the dissociation of the given compound and the concentration of its ions in water. In this case, we are given a concentration of 0.1 M NH4⁺ (ammonium ion) in a 100 mL solution.
First, we need to understand the dissociation of NH4⁺ in water. Ammonium ion acts as a weak acid and can donate a proton (H⁺). When it dissociates, it forms NH3 (ammonia) and H⁺ ions.
The dissociation reaction is as follows:
NH4⁺(aq) ⇌ NH3(aq) + H⁺(aq)
Since NH4⁺ is a weak acid, we can assume that the dissociation is not complete, and we will need to use the equilibrium constant expression (Ka) to calculate the concentration of H⁺ ions.
The Ka expression for NH4⁺ is:
Ka = [NH3][H⁺] / [NH4⁺]
We can assume that the [NH3] formed from the dissociation is very small compared to the initial concentration of NH4⁺; thus, we can consider it negligible. Therefore, the Ka expression simplifies to:
Ka = [H⁺][NH3] / [NH4⁺]
Next, we need the value of Ka for NH4⁺. The Ka value of NH4⁺ is 5.6 x 10⁻¹⁰ at 25°C.
Now, we can use the equation for the Ka expression to find the concentration of H⁺ ions. Rearranging the equation gives us:
[H⁺] = (Ka * [NH4⁺]) / [NH3]
Substituting the given values:
[H⁺] = (5.6 x 10⁻¹⁰ * 0.1 M) / 0 M
[H⁺] = 5.6 x 10⁻¹¹ M
Since pH is defined as the negative logarithm (base 10) of the H⁺ ion concentration, we can calculate the pH using the [H⁺] value we obtained:
pH = -log10([H⁺])
pH = -log10(5.6 x 10⁻¹¹)
pH ≈ 10.2
Therefore, the pH of the 0.1 M NH4⁺ solution is approximately 10.2.