Question

Chemistry Question : Solving for the molarity of an acid with the base's molarity already given. Help please?

If 75 ml of H2SO4 is neutralized by 160 ml of .75 moles of HOH what is the concentration (M) of the acid

Answer 1

0.80 mol/l

Explanation:

I'll assume you mean NaOH since HOH is water. I will also assume that you mean 0.75 M of NaOH.

Start with the equation:

`sf(H_2SO_4+2NaOHrarrNa_(2)SO_4+2H_2O)`

This tells us that 1 mole of `sf(H_2SO_4)` reacts with 2 moles of `sf(NaOH)`

Now `sf(c=n/v)`

`:.` `sf(n=cxxv)`

`:.` `sf(n_(NaOH)=0.75xx160/1000=0.12)`

The equation tells us that the no. of moles of `sf(H_2SO_4)` must be 1/2 of this so:

`sf(n_(H_2SO_4)=0.12/2=0.06)`

(Note that 75 ml = 75/1000 = 0.075 L)

Since `sf(c=n/v)` we can say that:

`sf([H_2SO_4]=0.06/0.075=0.80color(white)(x)"mol/l")`