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

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")#