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

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

Answer 2

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* 0.220M* 11.9ml=1* M_2* 25.0ml`

`M_2=0.209M`

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