What is the pH of a 0.150 M NH4CN
(H^+) = sqrt(KwKa/Kb), then convert to pH.
is the Kw 0.150M?
Kw = 1E-14
Ka is Ka for HCN
Kb is Kb for NH3.
To determine the pH of a solution, we need to consider the dissociation of the solute and the resultant concentration of hydrogen ions (H+). In the case of NH4CN, it is an acidic salt and undergoes hydrolysis to produce NH4+ and CN- ions:
NH4CN (aq) ⇌ NH4+ (aq) + CN- (aq)
The NH4+ ion acts as an acid by donating a proton (H+), while the CN- ion acts as a base by accepting a proton (H+). Therefore, we need to compare the acidity and basicity of NH4+ and CN- ions to determine the overall pH of the solution.
First, we need to determine whether NH4+ or CN- ionizes more strongly. NH4+ is a weak acid, whereas CN- is the conjugate base of a weak acid, making it a weak base. Comparing these two species, weak acids are stronger than their conjugate bases. Therefore, NH4+ ionizes more strongly compared to CN-.
Since NH4+ is the stronger acid, it will donate H+ ions to the solution, resulting in an acidic pH. To calculate the pH of the solution, we need to consider the concentration of NH4+ ions.
Given that the concentration of NH4CN is 0.150 M, and considering the stoichiometry of the dissociation, the concentration of NH4+ and CN- ions will also be 0.150 M.
Next, we will assume that the concentration of NH4+ ions dissolving will be negligible compared to the initial concentration of NH4+ (0.150 M). This approximation can be made due to the weak acidity of this ion.
Now, we need to calculate the concentration of H+ ions (protons) resulting from the ionization of NH4+ in water. The concentration of H+ ions can be determined using the equilibrium expression for NH4+ ionization:
Ka = [NH4+][H+]/[NH4CN]
For NH4CN, the Ka (acid dissociation constant) can be found in chemical reference sources. Its value is approximately 5.6 x 10^-10 at room temperature.
Using the Ka value and the concentration of NH4CN (0.150 M), we can set up the equation
5.6 x 10^-10 = [NH4+][H+] / 0.150
Rearranging the equation to solve for [H+], we get:
[H+] = (5.6 x 10^-10) * 0.150 / [NH4+]
Since we previously assumed that the concentration of NH4+ ions resulting from the dissociation of NH4CN is negligible compared to the initial concentration of NH4+ (0.150 M), we can consider [NH4+] as approximately 0.150 M.
Plugging in the values, we find:
[H+] = (5.6 x 10^-10) * 0.150 / 0.150
[H+] = 5.6 x 10^-10
Therefore, the concentration of H+ ions (protons) resulting from the hydrolysis of NH4CN is approximately 5.6 x 10^-10 M.
Finally, to calculate the pH, we convert the concentration of H+ ions to pH using the pH equation:
pH = -log[H+]
pH = -log(5.6 x 10^-10)
Calculating this value, we find that the pH of a 0.150 M NH4CN solution is approximately 9.25.