Calculate the pH of the following aqueous solution:
1.00 mol/L sulfuric acid, H2SO4(aq)
H2SO4 ==> H^+ + HSO4^- 100%
HSO4^- ==> H^+ + SO4^= not 100%
(H^+) = 1.00 M for the first ionization.
The second one is guided by k2.
k2 = (H^+)(HSO4^-)/(HSO4^-)
Plug into k2 as follows:
(H^+) = 1.00 + x
(SO4^=) = x
(HSO4^-) = 1.00 - x
Solve for x.
k2 = (H^+)(HSO4^-)/(HSO4^-)
= (1.00 + x)(1.00 - x)/(1.00 -x)
=Im guessing the 1.00 cancels out?
.
16 years ago
16 years ago
If you work the quadratic (it can be done with successive approximations), the answer comes out to be (H^+) = 1.1072 which I would round to 1.11 M and take - log that for pH of about 1.97.
1 year ago
Yes, you're correct. The 1.00 in the expression cancels out, leaving us with:
k2 = (1.00 + x)(1.00 - x)/(1.00 - x)
To solve for x, we can multiply both sides of the equation by (1.00 - x):
k2(1.00 - x) = (1.00 + x)(1.00 - x)
Expanding both sides of the equation:
k2 - k2x = 1.00 - x^2
Rearranging the equation:
x^2 - k2x + (k2 - 1.00) = 0
This is a quadratic equation in terms of x. We can solve it by using the quadratic formula:
x = (-b ± √(b^2 - 4ac))/(2a)
In this case, a = 1, b = -k2, and c = (k2 - 1.00). Plug in the values and solve for x.
11 months ago
To calculate the pH of the given aqueous solution (1.00 mol/L sulfuric acid, H2SO4(aq)), we need to consider the ionization of sulfuric acid.
Sulfuric acid (H2SO4) undergoes two ionization reactions:
1. H2SO4 → H+ + HSO4- (100% ionization)
2. HSO4- → H+ + SO42- (not 100% ionization)
First, we can calculate the concentration of H+ ions in solution after the first ionization. According to the equation, the concentration of H+ is equal to the concentration of the initial sulfuric acid solution, which is 1.00 M.
Now, for the second ionization, we can write a balanced equilibrium expression:
k2 = [H+][HSO4-] / [SO42-]
We assume that the value of k2 is known. In the calculation, we represent the unknown concentration of H+ as x and assume that it is small compared to 1.00 M. Thus, we can approximate:
[H+] = 1.00 M + x
[HSO4-] = 1.00 M - x
[SO42-] = x
Substituting these values into the equilibrium expression, we have:
k2 = (1.00 M + x)(1.00 M - x) / x
Since the concentration of H+ ions in the second ionization is small compared to 1.00 M, the term "1.00" can be ignored in the expression:
k2 = x(1.00 M - x) / x
k2 = 1.00 M - x
Now, we can solve for x by setting k2 equal to 1.00 M - x:
1.00 M - x = 1.00 M - x
0 = 0
This equation shows that x can have any value. Therefore, the second ionization of H2SO4 is not relevant to the calculation of the pH.
Hence, the pH of the 1.00 mol/L sulfuric acid solution is simply determined by the concentration of H+ ions after the first ionization, which is 1.00 M. The pH is given by the negative logarithm (base 10) of the H+ concentration:
pH = -log10(1.00) = 0
Therefore, the pH of the given solution is 0.