Calculate the pH of a 5.90 10-3 M solution of H2SO4
I have tried doing this as pH=-log[H+] and also by multiplying the Molarity by two because there are two hydrogens and then doing the -log but both ways have gotten me the wrong answer is there something i am missing?
same thing happened with: Calculate the pH of a 0.44 M solution of NaHSO4
please help, so confused
First, do you have answers. The first H^+ from H2SO4 is 100% ionized, the second one is not; therefore, the H^+ is not twice M H2SO4 but it is more than one times that. Second thing is the NaHSO4. That is not the M HSO4^- because
HSO4^- ==> H^+ + SO4^2- and k2 = about 0.012. If you have answers I can see how you are expected to solve this.
I worked this and I expect you are supposed to take all of this into consideration. Here is how you do it.
...........H2SO4 ==> H^+ + HSO4^-
initial....0.0059....0.......0
change....-0.0059..0.0059..0.0059
equil.......0......0.0059..0.0059
So the first ionization, which is 100%, produces 0.0059M H^+. Then the second H, which is not 100% ionized, kicks in.
............HSO4^- ==> H^+ + SO4^2-
initial....0.0059......0......0
change......-x.........x.......x
equil......0.0059-x....x.......x
k2 = (H^+)(SO4^2-)/(HSO4^-)
You need to use the k2 in your text or notes for they change from text to text. My copy I use list it as 0.012.
For (H^+) you want to substitute 0.0059+x (that's 0.0059 from the first H and x from the HSO4^-. For SO4^2- substitute x and for HSO4^- substitute 0.0059-x. This is a quadratic and you solve for x. Then total H^+ = 0.0059+x and determine pH from that.
For the NaHSO4, follow the second part of the above.
...........HSO4^- ==> H^+ + SO4^2-
initial...0.44.......0.......0
change......-x.......x........x
equil....0.44-x.......x.......x
k2 = (H^+)(SO4^2-)/(HSO4^-)
Substitute from the ICE chart and solve for x, then convert to pH.
To calculate the pH of a solution of H2SO4 or NaHSO4, you need to consider their ionization in water.
H2SO4 is a strong acid that completely dissociates into H+ and SO4^2- ions in water. Since there are two acidic hydrogens per molecule of H2SO4, we can assume that all of the H+ ions come from the dissociation of H2SO4.
NaHSO4 is a weak acid that only partially dissociates into H+ and HSO4^- ions in water. Since it is a weak acid, we need to consider the extent of ionization.
Let's calculate the pH of each solution step-by-step:
1. H2SO4:
The concentration of H+ ions in a 5.90 * 10^-3 M solution of H2SO4 is given by its molarity.
[H+] = 5.90 * 10^-3 M
Now, calculate the pH using the equation:
pH = -log[H+]
pH = -log(5.90 * 10^-3)
pH = -(-2.23) (approximately)
pH ≈ 2.23
Therefore, the pH of a 5.90 * 10^-3 M solution of H2SO4 is approximately 2.23.
2. NaHSO4:
Since NaHSO4 is a weak acid, it partially ionizes in water according to the equation: NaHSO4 ⇌ H+ + HSO4^-
The H+ concentration can be determined from the initial concentration of NaHSO4.
Given that the concentration of NaHSO4 is 0.44 M, we can assume that some proportion of it will dissociate into H+ and HSO4^- ions. For simplicity, let's assume that x is the concentration of H+ ions.
Initially, [HSO4^-] = [NaHSO4] = 0.44 M
At equilibrium, [H+] = x
The equilibrium expression for the dissociation of NaHSO4 is:
Ka = [H+][HSO4^-] / [NaHSO4]
Using the initial concentrations, the expression can be rearranged as:
Ka = x * (0.44 - x) / 0.44
Since Ka for NaHSO4 is given, you can solve this equation to find the value of x, which corresponds to [H+]. Once you find x, you can calculate the pH using the equation:
pH = -log[H+]
Unfortunately, you did not provide the value of Ka for NaHSO4, so I cannot give you the exact pH in this case.
To calculate the pH of a solution, you need to find the concentration of the hydrogen ions ([H+]) in the solution.
For the first question, you have a 5.90 × 10^-3 M solution of H2SO4. Since H2SO4 is a strong acid, it dissociates completely in water, releasing two H+ ions for every molecule of H2SO4. Therefore, the concentration of H+ ions in the solution is twice the molarity of H2SO4.
Concentration of H+ ions = 2 × 5.90 × 10^-3 M = 1.18 × 10^-2 M
Now that you have the concentration of H+ ions, you can find the pH using the equation: pH = -log[H+]
pH = -log(1.18 × 10^-2) ≈ 1.93
For the second question, you have a 0.44 M solution of NaHSO4. This is a weak acid, so it does not completely dissociate in water. However, since it is a monoprotic acid (releases only one H+ ion), the concentration of H+ ions is the same as the molarity of NaHSO4.
Concentration of H+ ions = 0.44 M
pH = -log(0.44) ≈ 0.35
So, the pH of a 5.90 × 10^-3 M solution of H2SO4 is approximately 1.93, and the pH of a 0.44 M solution of NaHSO4 is approximately 0.35.