Answer: 2.4 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]Hl[/tex] solution = 3.5 M
[tex]V_1[/tex] = volume of [tex]HCl[/tex] solution = 34.3 ml
[tex]M_2[/tex] = molarity of [tex]NaOH[/tex] solution = ?
[tex]V_2[/tex] = volume of [tex]NaOH[/tex] solution = 50 ml
[tex]n_1[/tex] = valency of [tex]HCl[/tex] = 1
[tex]n_2[/tex] = valency of [tex]NaOH[/tex] = 1
[tex]1\times 3.5M\times 34.3=1\times M_2\times 50[/tex]
[tex]M_2=2.4[/tex]
Therefore, the concentration of NaOH solution is 2.4 M.