Answer: The concentration of the [tex]NH_3[/tex] in solution is 1.42 M
Explanation:
According to neutralization law:
[tex]n_1M_1V_1=n_2M_2V_2[/tex]
where,
[tex]n_1[/tex] = basicity of [tex]HCl[/tex] = 1
[tex]n_2[/tex] = acidity of [tex]NH_3[/tex] = 1
[tex]M_1[/tex] = concentration of [tex]HCl[/tex] = 0.5000 M
[tex]M_2[/tex] = concentration of [tex]NH_3[/tex] = ?
[tex]V_1[/tex] = volume of [tex]HCl[/tex] = 42.6 ml
[tex]V_2[/tex] = volume of [tex]NH_3[/tex] = 15.00 ml
Now put all the given values in the above law, we get:
[tex](1\times 0.5000\times 42.6)=(1\times M_1\times 15.00)[/tex]
By solving the terms, we get :
[tex]M_1=1.42M[/tex]
Thus the concentration of the [tex]NH_3[/tex] in solution is 1.42 M