Question

If 26.35 ml of a standard 0.1650 m naoh solution is required to neutralize 35.00 ml of h2so4, what is the molarity of the acid solution?

Answer 1

 26.35 ml NaOH x (.1650 moles NaOH / 1000 ml ) x (1 mole H2SO4 / 2 moles NaOH) = moles H2SO4 = .002174 moles H2SO4 

molarity of H2SO4 = .002174 / .035 L = .06211 M

Answer 2

Answer : The molarity of the $H_2SO_4$ is, 0.06211 M

Explanation :

Using neutralization law,

$n_1M_1V_1=n_2M_2V_2$

where,

$n_1$ = basicity of an acid $(H_2SO_4)$ = 2

$n_2$ = acidity of a base $(NaOH)$ = 1

$M_1$ = concentration or molarity of $H_2SO_4$ = ?

$M_2$ = concentration of NaOH = 0.1650 M

$V_1$ = volume of $H_2SO_4$ = 35.00 mL

$V_2$ = volume of NaOH = 26.35 mL

Now put all the given values in the above law, we get the concentration of the $H_2SO_4$.

$2\times M_1\times 35.00mL=1\times 0.1650M\times 26.35mL$

$M_1=0.06211M$

Therefore, the molarity of the $H_2SO_4$ is, 0.06211 M