Question

Automobile batteries use 3.0 m h2so4 as an electrolyte. how much 1.20 m naoh will be needed to neutralize 225 ml of battery acid? h2so4(aq) + 2naoh(aq) --> 2h2o(l) + na2so4(aq)

Answer 1

Answer : The volume of the
`NaOH` needed are, 1125 ml

Explanation :

Using neutralization law,

`n_1M_1V_1=n_2M_2V_2`

where,

`n_1` = basicity of an acid = 2

`n_2` = acidity of a base = 1

`M_1` = concentration of
`H_2SO_4` = 3.0 M

`M_2` = concentration of NaOH = 1.20 M

`V_1` = volume of
`H_2SO_4` = 225 ml

`V_2` = volume of NaOH = ?

Now put all the given values in the above law, we get the volume of the
`NaOH`.

`2* 3.0M* 225ml=1* 1.20M* V_2`

`V_2=1125ml`

Therefore, the volume of the
`NaOH` needed are, 1125 ml

Answer 2

H2SO4 (aq) + 2 NaOh (aq) ------>2H2O + Na2SO4(aq)

V1M1 n2= V2M2n1
V= volume
M= concentration in mole per liter
n= number of moles
V1=?
V2= 225 ml
M1= 1.2 M
M2= 3 m
n1=2 moles
V1 is therefore = ( 225 x3 x2 ) /1.2 = 1125 ml