I’m trying to find the molarity of h2so4 in this calculation. i attempted it but i still need someone to check my work. i used 10.00 ml of h2so4 and the volume added to it was 24.97 ml of the 0.0924m naoh solution. the equation is 2naoh + h2so4 -> 2h2o + na2so4, so i’m aware 1 mol of h2so4 is 2 mol naoh. what i tried: 0.0924 mol/l x 0.02497 l= 0.002307 mol naoh x 1 mol h2so4/2 mol naoh= 0.001156 mol h2so4 0.0011536 mol h2so4/0.01000 l h2so4= 0.1154m h2so4 am i doing this right?

Since it goes to completion, we can use:  

Use NaMaVa =NbMbVb  

where  

Na= moles of H+ in H2SO4  

Ma= molarity of the acid  

Va= volume of the acid  

Nb= moles of OH- in NaOH  

Mb= molarity of the base  

Vb= molarity of the base  

plug in the numbers  

(2 moles of H+) (Volume acid) (10.00mL acid) = (1 mole of OH-) (.248M base) (18.71mL base)  

20 V = 4.64  

V= .232 M

a

explanation:

the atomic number will stay the same no matter how many electrons it has, so it doesn’t decrease nor increase. it’ll only change with a different number of protons