Consider a 0.80M solution of HCN. The Ka of HCN is 6.2x10-10. What are the major species in solution? Calculate the pH of this solution.
..........HCN ==> H^+ + CN^-
initial...0.80M...0......0
change.....-x......x.....x
equil....0.80-x....x.....x
Ka = (H^+)(CN^-)/(HCN)
Substitute and solve for x and convert to pH.
After you finish the calculation you will know what the major species are. Don't forget H2O as a major component.
To determine the major species in a solution of HCN and calculate the pH, we need to consider the dissociation of HCN in water.
The dissociation of HCN in water can be represented as follows:
HCN + H2O ⇌ H3O+ + CN-
From the balanced equation, we can see that the major species in solution are H3O+ (hydronium ion) and CN- (cyanide ion).
To calculate the pH of the solution, we need to find the concentration of H3O+ ions.
Since the initial concentration of HCN is given as 0.80 M, we can assume that the concentration of HCN after dissociation is x M.
The equilibrium expression for the dissociation reaction of HCN can be written as follows:
Ka = [H3O+][CN-] / [HCN]
Given that the Ka of HCN is 6.2x10^-10, we can substitute the known values into the equation:
6.2x10^-10 = [H3O+][CN-] / x
Assuming x ≈ [H3O+], we can rewrite the equation as:
6.2x10^-10 = [H3O+]^2 / 0.80
Next, we can rearrange and solve the equation for [H3O+]:
[H3O+]^2 = 0.80 * 6.2x10^-10
[H3O+]^2 = 4.96x10^-10
[H3O+] ≈ √(4.96x10^-10)
[H3O+] ≈ 7.03x10^-6 M
Finally, to calculate the pH of the solution, we can use the relationship:
pH = -log[H3O+]
pH = -log(7.03x10^-6)
pH ≈ 5.15
Therefore, the major species in the solution are H3O+ (hydronium ion) and CN- (cyanide ion), and the pH of the solution is approximately 5.15.
To determine the major species in the solution, we need to consider the dissociation of HCN in water. HCN is a weak acid, and it will partially dissociate to form its conjugate base, CN-, and a few H+ ions.
The balanced chemical equation for the dissociation of HCN in water can be written as follows:
HCN + H2O ⇌ CN- + H3O+
Since the solution concentration of HCN is 0.80 M, we can assume that x moles of HCN dissociate. Therefore, the concentration of HCN remaining in the solution will be (0.80 - x) M.
Given the Ka value of HCN as 6.2x10-10, we can use an ICE table (initial, change, equilibrium) to find the concentration of CN- and H3O+ ions at equilibrium.
The Ka expression for this reaction is:
Ka = [CN-][H3O+]/[HCN]
At equilibrium, the HCN concentration is (0.80 - x) M, and the CN- and H3O+ concentrations are both x M. Plugging these values into the Ka expression, we get:
Ka = (x)(x)/(0.80 - x)
Next, we can assume that x is small compared to 0.80, so we can approximate (0.80 - x) as 0.80:
Ka = (x)(x)/(0.80)
Simplifying further, we obtain:
6.2x10-10 = x^2/(0.80)
Rearranging the equation:
x^2 = (6.2x10-10)(0.80)
x^2 = 4.96x10-10
x = √(4.96x10-10)
x ≈ 7.04x10-6 M
Now we know that the concentration of CN- and H3O+ ions in the solution is approximately 7.04x10-6 M.
To calculate the pH of the solution, we need to determine the concentration of H+ ions. Since H+ and H3O+ ions are essentially the same, the concentration of H+ ions is also 7.04x10-6 M.
Using the formula pH = -log[H+], we can now calculate the pH:
pH = -log(7.04x10-6)
pH ≈ 5.15
Therefore, the major species in the solution are CN- and H3O+ ions, and the pH of the solution is approximately 5.15.