Answer: The concentration of [tex]H_2SO_4[/tex] is 0.234 M
Explanation:
According to the neutralization law,
[tex]n_1M_1V_1=n_2M_2V_2[/tex]
where,
[tex]n_1[/tex] = basicity [tex]H_2SO_4[/tex] = 2
[tex]M_1[/tex] = molarity of [tex]H_2SO_4[/tex] solution = ?
[tex]V_1[/tex] = volume of [tex]H_2SO_4[/tex] solution = 50.0 ml
[tex]n_2[/tex] = acidity of [tex]NaOH[/tex] = 1
[tex]M_1[/tex] = molarity of [tex]NaOH[/tex] solution = 0.375 M
[tex]V_1[/tex] = volume of [tex]NaOH[/tex] solution = 62.5 ml
Putting in the values we get:
[tex]2\times M_1\times 50.0=1\times 0.375\times 62.5[/tex]
[tex]M_1=0.234M[/tex]
Therefore concentration of [tex]H_2SO_4[/tex] is 0.234 M