Calculate the pH of the following aqueous solution:
1.00 mol/L sulfuric acid, H2SO4(aq)
H2SO4 is one of those acids that is 100% ionized for the first H^+ and less than 100% for the second one.
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.
Post your work if you get stuck. This is a quadratic you must solve BUT it may be easier to use approximations. You will need to look up k2 for H2SO4.
To calculate the pH of an aqueous solution, we need to know the concentration of hydrogen ions (H+). In the case of sulfuric acid (H2SO4), it's a strong acid that dissociates completely in water, producing two hydrogen ions for each molecule of acid.
So, for a 1.00 mol/L solution of sulfuric acid (H2SO4), the concentration of hydrogen ions (H+) is also 1.00 mol/L.
The pH scale is a logarithmic scale that measures the acidity or basicity of a solution. The pH is defined as the negative logarithm (base 10) of the hydrogen ion concentration:
pH = -log[H+]
Therefore, to calculate the pH of the sulfuric acid solution, we can take the logarithm of 1.00 and then multiply by -1:
pH = -log(1.00) = -(-0.00) = 0.00
So, the pH of a 1.00 mol/L sulfuric acid solution is 0.00.