Question

What is `pH` for a `1.95*g` mass of pure sulfuric acid added to enough water to give a final volume of `1*dm^3`?

Answer 1

`pH=1.40........`

Explanation:

We assume complete dissociation of the sulfuric acid:

`H_2SO_4(aq) + 2H_2O(l) rarr 2H_3O^+ + SO_4^(2-)`

And so `[H_2SO_4]` `-=` `((1.95*g)/(98.08*g*mol^-1))/(1*dm^3)` `=` `0.0199*mol*L^-1`. However, this is a FORMAL concentration and corresponds to `[SO_4^(2-)]`. `[H_3O^+]=0.0398*mol*L^-1`, i.e. TWICE the formal concentration.

Now since, `pH=-log_10[H_3O^+]`, `pH=-log_10(0.0398)=??`