Answer: The molarity of the borax solution is 0.107 M
Explanation:
The neutralization reaction is:
[tex]Na_2B_4O_7.10H_2O+H_2SO_4(aq)\rightarrow Na_2SO_4+4H_3BO_3+5H_2O[/tex]
According to neutralization law:
[tex]n_1M_1V_1=n_2M_2V_2[/tex]
where,
[tex]n_1[/tex] = basicity of [tex]H_2SO_4[/tex] = 2
[tex]n_2[/tex] = acidity of borax = 2
[tex]M_1[/tex] = concentration of [tex]H_2SO_4[/tex] = 1.03 M
[tex]M_2[/tex] = concentration of borax =?
[tex]V_1[/tex] = volume of [tex]H_2SO_4[/tex] = 2.07ml
[tex]V_2[/tex] = volume of borax = 20.0 ml
Now put all the given values in the above law, we get the molarity of borax:
[tex](2\times 1.03\times 2.07)=(2\times M_2\times 20.0)[/tex]
By solving the terms, we get :
[tex]M_2=0.107M[/tex]
Thus the molarity of the borax solution is 0.107 M
