Answer :
Answer: The concentration of [tex]NaOH[/tex] is 0.08 M.
Explanation:-
According to the neutralization law,
[tex]n_1M_1V_1=n_2M_2V_2[/tex]
where,
[tex]M_1[/tex] = molarity of [tex]HCl[/tex] solution = 0.105 M
[tex]V_1[/tex] = volume of [tex]HCl[/tex] solution = 25 ml
[tex]M_2[/tex] = molarity of [tex]NaOH[/tex] solution = ?
[tex]V_2[/tex] = volume of [tex]NaOH[/tex] solution = 31.5 ml
[tex]n_1[/tex] = valency of [tex]HCl[/tex] = 1
[tex]n_2[/tex] = valency of [tex]NaOH[/tex] = 1
[tex]1\times 0.105M\times 25=1\times M_2\times 31.5[/tex]
[tex]M_2=0.08M[/tex]
Therefore, the concentration of [tex]NaOH[/tex] is 0.08 M.