1)Calculate the pH of a 0.10 M NaHX salt solution. Ka (H2X) = 2.1 x 10-2 Ka (HX-) = 4.5 x 10-8
2)Calculate the pH of a 0.10 M NaHX salt solution. Ka (H2X) = 5.3 x 10-5 Ka (HX-) = 5.3 x 10-11
http://www.jiskha.com/display.cgi?id=1319241957
It helps us if you keep the same screen name. Posting under several names gets no extra help; in fact, it hinders it.
?? helps little. The first problem in your post is worked exactly like the link I gave you.
To calculate the pH of a salt solution, we need to determine whether the salt is an acidic, basic, or neutral salt. This can be done by examining the acid-base properties of the anion and cation present in the salt.
1) In the case of NaHX,
The anion X- comes from the acid HX-. The acid HX- has a Ka value of 4.5 x 10^-8, indicating that it is a weak acid. Therefore, we can consider the X- ion as a weak base.
The cation Na+ comes from the strong base NaOH, which is considered neutral.
Since the anion (X-) is a weak base and the cation (Na+) is neutral, the salt NaHX will create a slightly acidic solution.
To calculate the pH:
We can make the assumption that the concentration of X- is negligible compared to the concentration of HX-, as HX- is a weak acid.
Therefore, we can consider the dissociation of NaHX as follows:
NaHX → Na+ + HX-
The HX- will react with water in a hydrolysis reaction:
HX- + H2O ↔ H2X + OH-
We can now create an equilibrium expression:
Ka = [H2X][OH-] / [HX-]
Since the concentration of water remains relatively constant, we can ignore it in our calculations. Therefore, the expression simplifies to:
Ka = [H2X][OH-] / [HX-]
We know that [HX-] is equal to the initial concentration of NaHX, which is 0.10 M.
We can assign [OH-] as x, as it will be a small value in comparison to the concentration of HX-.
[H2X] can also be assigned as x.
Now, we can plug these values into the equation:
Ka = x * x / 0.10
Solve for x by rearranging the equation:
x^2 = 0.10 * Ka
x = √(0.10 * Ka)
Calculate the concentration of H+ using the equation: [H+] = 10^(-pH).
Therefore, pH = -log[H+].
Finally, substitute the concentration of H+ into the pH equation to find the pH of the solution.
2) The second question follows the same steps as the first one.
The anion X- comes from the acid HX-, which has a Ka value of 5.3 x 10^-11. Again, this indicates that it is a weak acid.
Since the anion is a weak acid and the cation (Na+) is neutral, the salt NaHX will also create a slightly acidic solution.
Follow the same steps as in the first question to calculate the pH. Plug in the given Ka values, solve for x, and then calculate the pH using the concentration of H+.