Question

What is the molarity of a naoh solution if 11.9 ml of a 0.220 m h2so4 solution is required to neutralize a 25.0-ml sample of the naoh solution?

Answer 1

Answer : The concentration of the $NaOH$ is, 0.209 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 $H_2SO_4$ = 0.220 M

$M_2$ = concentration or molarity of NaOH = ?

$V_1$ = volume of $H_2SO_4$ = 11.9 ml

$V_2$ = volume of NaOH = 25.0 ml

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

$2\times 0.220M\times 11.9ml=1\times M_2\times 25.0ml$

$M_2=0.209M$

Therefore, the concentration of the $NaOH$ is, 0.209 M

Answer 2

0.355M x 0.0282L= 0.01 moles of H2SO4. Remember sulphuric acid is diprotic so it will release 2 from each molecule. 
So moles of protons = 0.01 x 2 = 0.02 moles of H+ 
For neutralization: moles H+ = moles OH- 
Therefore moles of NaOH = 0.02 
conc = moles / volume 
Conc NaOH = 0.02 / 0.025L = 0.8M