Calculate the pH of a 2.67×10-3 M solution of H2SO4.
(Ka = 0.0120 for HSO4-)
H2SO4 is a diprotic acid. The first H is ionized completely but the second one has a k2 = 0.0120. We need to go through two ICE charts.
For the first H^+ it is as follows:
.......................H2SO4 ==> H^+ + HSO4^-
I...................0.00267...........0...........0
C.................-0.00267......0.00267....0.00267
E........................0..........0.00267.....0.00267 so the first H^+ contributes 0.00267 M. The second H is done this way.
....................HSO4^- ==> H^+ + SO4^2-
I..................0.00267........0.00267.....0
C..................-x..................+x.............x
E............0.00267 - x.......0.00267+x...x
Then k2 = (H^+)(SO4^2-)/(HSO4^-)
Substitute the E line into k2 expression and solve for x (you will need to solve the quadratic equation), evaluate 0.00267 + x which will give you the total (H^+), then convert that to pH. Post your work if you get stuck.
To calculate the pH of the solution, we need to take into account the Ka value for HSO4- and the concentration of the H2SO4 solution. We can use the equation for Ka to find the concentration of the H+ ions in the solution, and then calculate the pH from there.
Step 1: Write the balanced equation for the dissociation of H2SO4:
H2SO4 ⇌ H+ + HSO4-
Step 2: Write the expression for the equilibrium constant Ka:
Ka = [H+][HSO4-] / [H2SO4]
We are given the value for Ka (0.0120), so we can rearrange the equation to solve for [H+]:
[H+] = (Ka * [H2SO4]) / [HSO4-]
Step 3: Substitute the values into the equation:
[H+] = (0.0120 * 2.67×10-3) / (2.67×10-3)
[H+] = 0.0120
Step 4: Finally, calculate the pH using the formula:
pH = -log[H+]
pH = -log(0.0120)
pH ≈ 1.92
Therefore, the pH of the 2.67×10-3 M solution of H2SO4 is approximately 1.92.