Answer: The molarity of [tex]Ba(OH)_2[/tex] is 0.07 M
Explanation:
According to the neutralization law,
[tex]n_1M_1V_1=n_2M_2V_2[/tex]
where,
[tex]n_1[/tex] = basicity [tex]HCl[/tex] = 1
[tex]M_1[/tex] = molarity of [tex]HCl[/tex] solution = 0.1 M
[tex]V_1[/tex] = volume of [tex]HCl[/tex] solution = 35 ml
[tex]n_2[/tex] = acidity of [tex]Ba(OH)_2[/tex] = 2
[tex]M_2[/tex] = molarity of [tex]Ba(OH)_2[/tex] solution = ?
[tex]V_2[/tex] = volume of [tex]Ba(OH)_2[/tex] solution = 25 ml
Putting in the values we get:
[tex]1\times 0.1\times 35=2\times M_2\times 25[/tex]
[tex]M_2=0.07[/tex]
Therefore, molarity of [tex]Ba(OH)_2[/tex] is 0.07 M
