1. Determine the pH of a solution containing 1.00 mol/L of HCN (Ka = 5.00×10–10) and 5.00×10–2 mol/L of HCl.
Essentially one ignores the contribution of H^+ from the HCN but you can calculate how much that is.
.......HCN ==> H^+ + CN^-
I......1........0.....0
C......-x.......x.....x
E......1-x......x.....x
For HCl, remember it is a strong electrolyte and ionizes completely.
........HCl ==> H^+ + Cl^-
I.....0.05......0......0
C....-0.05....0.05....0.05
E........0....0.05....0.05
E....
Ka for HCN = (H^+)(CN^-)/(HCN)
(H^+) = 0.05+x (that's 0.05 from the HCl and x from the HCN)
HCN is 1-x. Solve for (CN^-) and that = (H^+) from the HCN.
Compare with that from the HCl. You will see it is negligible.
Then pH = -log(HCl) = ?
P.S. Note the correct spelling of chemistry.
To determine the pH of a solution containing both HCN and HCl, we need to consider the individual acid dissociations and calculate the concentration of H+ ions.
First, we'll start with HCN (a weak acid) and HCl (a strong acid). Since HCl is a strong acid, it dissociates completely in solution, and its concentration of H+ ions will be equal to its initial concentration.
For HCN, we need to consider its acid dissociation using the given Ka value. The Ka expression for HCN is:
HCN(aq) ⇌ H+(aq) + CN-(aq)
The equilibrium expression is:
Ka = [H+][CN-] / [HCN]
Given that the initial concentration of HCN is 1.00 mol/L and the Ka value is 5.00×10^–10, we can say that x is the concentration of H+ ions and CN- ions. So, at equilibrium, the concentration of HCN will be 1.00 - x.
Since x is negligible compared to 1.00, we can approximate 1.00 - x to be 1.00. Therefore, the concentration of HCN at equilibrium is 1.00 mol/L.
Now, the equilibrium expression becomes:
5.00×10^–10 = x^2 / 1.00
Solving for x, we find that x = √(1.00 * 5.00×10^–10) = 7.07×10^–6.
So, the concentration of H+ ions from HCN is 7.07×10^–6 mol/L.
For HCl, since it's a strong acid, its concentration of H+ ions is the same as its initial concentration, which is 5.00×10^–2 mol/L.
Now, we can calculate the total concentration of H+ ions in the solution:
[H+]total = [H+]from HCN + [H+]from HCl
[H+]total = 7.07×10^–6 mol/L + 5.00×10^–2 mol/L
[H+]total = 5.00×10^–2 mol/L + 7.07×10^–6 mol/L
[H+]total = 5.000000707 mol/L
Finally, we can calculate the pH of the solution using the formula:
pH = -log[H+]
pH = -log(5.000000707)
pH ≈ 1.30
Therefore, the pH of the solution containing 1.00 mol/L of HCN and 5.00×10^–2 mol/L of HCl is approximately 1.30.