Calculate the pH of a 2.00 M NH4CN solution.
The easy way to do this is to remember a formula I had in school many many years go.
(H^+) = sqrt (KwKa/Kb) or
pH =1/2pKw + 1/2pKa - 1/2pKb
To calculate the pH of a solution, we need to know the concentration of hydronium ions (H3O+). NH4CN is a weak acid, meaning it will partially ionize in water. The dissociation reaction of NH4CN is as follows:
NH4CN → NH4+ + CN-
Since NH4CN is a weak acid, we can assume that the concentration of H3O+ ions resulting from the dissociation of NH4CN is negligible. Therefore, to calculate the pH, we need to determine the concentration of hydroxide ions (OH-) in the solution.
To find the concentration of OH- ions, we need to use the equation for the dissociation of water:
H2O ↔ H+ + OH-
In pure water, the concentration of H+ ions is equal to the concentration of OH- ions, which is 1.0 x 10^(-7) M at 25°C.
However, since we are dealing with an NH4CN solution, we need to consider the hydrolysis of CN- ions with water:
CN- + H2O ↔ HCN + OH-
The concentration of CN- ions can be calculated using the initial concentration of NH4CN (2.00 M) and its dissociation constant (Ka). The Ka for NH4CN is 5.6 x 10^(-10) at 25°C.
Now, let's calculate the concentration of CN- ions:
[C(N)/C(NH4CN)] = √(Ka / [H(NH4CN)])
[C(N)/2.00] = √(5.6 x 10^(-10) / 2.00)
[C(N)] = 1.33 x 10^(-5) M
Next, let's calculate the concentration of OH- ions resulting from the hydrolysis reaction:
[OH-] = [HCN] = [C(N)] = 1.33 x 10^(-5) M
Since [OH-] = [H+], the pH is calculated as follows:
pH = -log([H+]) = -log(1.33 x 10^(-5)) = 4.88
Therefore, the pH of the 2.00 M NH4CN solution is approximately 4.88.