What is the pH ( 3 significant digits) of a 0.01 M solution of sulfuric acid? Hint is that the first ionization constant is "large", and the second is finite, and the [H+] donated by the second equilibrium depends upon the amount of [H+] already donated by the first ionization. What is the answer please.

For Further Reading

chemistry - DrBob222, Wednesday, March 26, 2008 at 1:29am
H2SO4 ==> H^+ + HSO4^-
HSO4^- ==> H^+ + SO4^-2

The first ionization is complete; i.e., it produces 0.01 M H^+.
The second ionization is not complete and obeys k2 = (H^+)(SO4^-)/(HSO4^-)

The easiest way to work the problem, I think, is to write the expression for k2, look up k2 for its value, then substitute as follows:
(H^+) = 0.01 + x
(SO4^-2) = x
(HSO4^-) = 0.01 - x
Then solve the quadratic for x.
Total (H^+) then is 0.01 + x.

The k2 value is 1.2 x 10^-2.. where exactly am I substituting that? can you show me how using that number.

For Further Reading

chemistry - Dr Russ, Wednesday, March 26, 2008 at 9:20am
DrBob has given you the equation to use:

k2 = [H^+][SO4^-]/[HSO4^-]

and the values to substitute

[H^+] = 0.01 + x
[SO4^-2] = x
[HSO4^-] = 0.01 - x
k2 = 1.2 x 10^-2

you need to substitute these and solve for x.

Remember that you are after 0.01+x [H^+]
to calculate the pH.

I got the answer of either 4.0 or 1.65.. but I don't know if it is right.. Can you refresh my memory on how to solve for x in this equation please.

To solve for x in the equation, k2 = [H+][SO4-]/[HSO4-], you can follow these steps:

1. Substitute the given values into the equation:
k2 = 1.2 x 10^-2
[H+] = 0.01 + x
[SO4^-2] = x
[HSO4^-] = 0.01 - x

2. Rewrite the equation with the substitutions:
1.2 x 10^-2 = (0.01 + x)(x) / (0.01 - x)

3. Simplify the equation by multiplying both sides by (0.01 - x):
1.2 x 10^-2 (0.01 - x) = (0.01 + x)(x)

4. Expand the equation:
0.012 - 0.012x = (0.01 + x)(x)

5. Distribute the terms on the right side:
0.012 - 0.012x = 0.01x + x^2

6. Rearrange the equation to set it equal to zero:
x^2 + 0.022x - 0.012 = 0

7. Solve the quadratic equation using the quadratic formula or factoring.

Using the quadratic formula:
x = (-0.022 ± sqrt(0.022^2 - 4×1×(-0.012))) / (2×1)
x = (-0.022 ± sqrt(0.000484 + 0.048)) / 2
x = (-0.022 ± sqrt(0.048484)) / 2

8. Calculating the values:
x ≈ (-0.022 + 0.219) / 2 ≈ 0.099
x ≈ (-0.022 - 0.219) / 2 ≈ -0.120

Since the concentration of ions cannot be negative, we discard the negative value.

9. Substitute the calculated value of x back into the expression for [H+]:
[H+] = 0.01 + x
[H+] ≈ 0.01 + 0.099
[H+] ≈ 0.109

10. Calculate the pH using the concentration of H+:
pH = -log[H+]
pH ≈ -log(0.109)
pH ≈ 1.96

Therefore, the pH of a 0.01 M solution of sulfuric acid is approximately 1.96 (to 3 significant digits).