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#?

1 Answer
Mar 25, 2017

Answer:

#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)=??#