Calculate the ph of 0.1M solution of NH4CN given that the dissociate constant Ka for HCN 0.0000000006 and Kb for NH3 is 0.000016
To calculate the pH of a solution of NH4CN, we need to consider the dissociation of the weak acid HCN and the weak base NH3.
Step 1: Write the balanced chemical equation for the dissociation of NH4CN.
NH4CN(aq) ⇌ NH4+(aq) + CN-(aq)
Step 2: Identify the acidic and basic species involved.
In this case, NH4+ is acidic and CN- is basic.
Step 3: Write the equilibrium expressions.
HCN ⇌ H+(aq) + CN-(aq)
NH3 + H2O ⇌ NH4+(aq) + OH-(aq)
The dissociation constant for HCN, Ka, is given as 0.0000000006. Therefore, we can write:
Ka = [H+][CN-] / [HCN]
0.0000000006 = [H+][CN-] / [HCN]
Now, we need to consider the reaction between NH4+ and H2O, which is the reverse of the reaction between NH3 and H2O:
NH4+(aq) + H2O(l) ⇌ NH3(aq) + H3O+(aq)
The dissociation constant for NH3, Kb, is given as 0.000016. Now, we can write the equilibrium expression:
Kb = [NH3][H3O+] / [NH4+]
0.000016 = [NH3][H3O+] / [NH4+]
Step 4: We can use the equilibrium expressions and the relationship between Ka and Kb to calculate the concentration of H+ ions in the solution and, consequently, the pH.
[H+][CN-] / [HCN] = 0.0000000006 (1)
[NH3][H3O+] / [NH4+] = 0.000016 (2)
However, we don't have the concentrations of HCN, CN-, NH3, H3O+, and NH4+ in this problem. We only have the initial concentration of the NH4CN solution, which is 0.1 M.
To solve this, we need to make assumptions:
1. Since NH4CN is a salt, it will completely dissociate into NH4+ and CN-. Therefore, the initial concentrations of NH4+ and CN- are both 0.1 M.
2. Since NH4CN is a weak acid, we can assume that the dissociation of HCN will be negligible compared to the initial concentration of NH4CN. Therefore, the initial concentration of HCN is approximately equal to the initial concentration of NH4CN, which is 0.1 M.
Using these assumptions, we can substitute the concentrations into the equilibrium expressions:
[H+][0.1] / [0.1] = 0.0000000006
[H+][CN-] = 0.0000000006
[NH3][H3O+] / [0.1] = 0.000016
[NH3][H3O+] = 0.0000016
Since the concentration of CN- is equal to the concentration of NH3 in this case, we can substitute the concentration of NH3 into equation (1):
[H+][NH3] = 0.0000000006
Multiplying the above two equations, we get:
[H+]^2 = (0.0000000006)(0.0000016)
[H+]^2 = 0.00000000000000000096
Taking the square root of both sides:
[H+] ≈ 0.00000000031
Finally, we can calculate the pH:
pH = -log[H+]
pH ≈ -log(0.00000000031)
pH ≈ 9.51
Therefore, the pH of the 0.1 M solution of NH4CN is approximately 9.51.