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.
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Ok, so i took a break, came back, and finished the problem. I have an x value of 2.23x10-5. I took the negative log of that and came up with a pH of 4.65, Is this close to what it should be, or did i miss something?
You didn't miss anything. That's the correct answer.
To determine the major species in the solution, you need to consider the ionization of HCN. HCN is a weak acid, so it will partially ionize in water. The ionization reaction of HCN in water can be represented as follows:
HCN (aq) ⇌ H+ (aq) + CN- (aq)
Since HCN is a weak acid, its major species in solution will predominantly be HCN and a smaller amount of H+ and CN- ions.
Now, let's move on to calculate the pH of the solution. The pH of a solution is a measure of its acidity and is related to the concentration of H+ ions present. To do this, we need to calculate the concentration of H+ ions in the solution.
Given that you have a 0.80 M solution of HCN, we can assume that the concentration of HCN is equal to the initial concentration (0.80 M). Let's call this initial concentration [HCN]0.
Since HCN partially ionizes, let x be the concentration (in M) of H+ (and CN-) ions formed in the solution. Therefore, [H+] and [CN-] are both equal to x.
Using the equilibrium constant expression for the ionization of HCN (Ka), we can write:
Ka = [H+][CN-] / [HCN]
Substituting the given values, Ka = 6.2x10^-10, [H+] = x, [CN-] = x, and [HCN] = 0.80 M, we can rearrange the expression as follows:
6.2x10^-10 = x * x / 0.80
Solving this quadratic equation, we find that x ≈ 2.23x10^-5 M.
Now, to calculate the pH, we take the negative log of the concentration of H+ ions:
pH = -log[H+]
Substituting the value of x in the equation, we get:
pH = -log(2.23x10^-5) ≈ 4.65
Therefore, your calculated pH of 4.65 is close to the expected value. You have successfully determined the major species in solution and calculated the pH of the solution accurately.