Question

calculate the PH of 0.25mol H2SO4
Ka= 1.9 x 10-5

only have experience with simpler ones like KOH, the H2SO4 is whats confusing me

Answer 1

if it doesnt state two Ka you consider it completely dissolves:
H2SO4 -> 2H+ + SO42-
and all you need to be careful with is concentration of H+ which is
Ka = [H+]^2 * [SO42-] / [H2SO4]
hope this helps ;)

Answer 2

Thanks thats exactly what I needed :P
this site is great

Answer 3

I think that u do the ice table
multiply Ka by the 0.25 take the square root which of is 2.18*10^-3
take the -log of that u get 2.66
to find the PH u subtract 14-2.66 and u get 11.34
Ph=11.34

Answer 4

Sulfuric acid is a polyprotic acid, meaning it can release two protons. Sulfuric acid itself is a strong acid, and dissociates completely in water solution like this:
\[{\rm H}_2{\rm SO}_4 (aq) \rightarrow {\rm H}^{+}(aq) + {\rm HSO}_{4}^{-}(aq)\]
The hydrogen sulfate that remains is a weak acid, and an equilibrium is reached for its dissociation:
\[{\rm HSO}_{4}^{-}(aq) \rightleftharpoons {\rm H}^{+}(aq) + {\rm SO}_{4}^{2-}(aq)\]
The Ka for this equilibrium is actually 1.2 x 10^-2, which is not the Ka you've got there....God knows what that is.

To solve for the pH of a polyprotic acid, take it in two steps. First, all of the H2SO4 dissociates, so initially [H+] = 0.25 M, [HSO4-] = 0.25 M. Now set up your equilibrium expression for the second step:
\[K_a = \frac{[{\rm H}^+][{\rm SO}_{4}^{2-}]}{{[\rm HSO}_{4}^{-}]}\]
Let x = the molarity of sulfate anion at equilibrium:
\[1.2 \times 10^{-2} = \frac{(0.25 + x)(x)}{0.25 - x}\]
Solving the quadratic gives x = 0.011 as the only plausible root. Hence the equilibrium concentration of [H+] is 0.261... and the pH = -0.58 to two significant decimal places (which is what you should report, since your data has two significant digits).

The result if you ignore the second dissociation, and just stop with the strong acid dissociation is [H+] = 0.25 M and pH = -0.60. So the second dissociation bumps up the pH very slightly.

This is a pretty sophisticated calculation if you're just used to strong acid and strong base pH calculations, so it's not clear to me you're expected to do all this. If you havent' dealt with weak acid equilibria at all it seems unlikely.

Answer 5

callandre300 and Carl_Pham i do recognize your answers but i believe that he wanted simple answer cause these are simple tasks... your answers may be right in tougher tasks but i believe he is solving simple task and not so comprehensive task ;)

Answer 6

What on earth does this Ka represent?
The question must have mixed sulfuric with something else.

Answer 7

Ka would be constant of acid and for bases it is Kb

Answer 8

Hi! I meant what does this value of acidity constant represent? As Carl_Pham said, there is no such value for H2SO4.
HSO4- is much more acidic than that.

Answer 9

well yes, but like i said, this is one of test tasks just to get in this type of calculations, it doesnt represent true picture of acidity but it is to get the grasp of calculating with Ka....

Answer 10

I dunno, K. Very few instructors would give a bogus physical constant just to get students used to the method of calculation. That would be like asking you to calculate wavelength from frequency, but giving you the wrong value for the speed of light. Who would do that?

It seems much more likely the question was garbled, or contained a mistake, in either the acid identity or the Ka.