Calculate the volume of 1.5 N ` H_2SO_4 ` completely neturalized by 35.8 mL of
` 2.5 N Ba(OH) _2`

Explanation

` 2.5N Ba (OH_2) `
Volume of `H_2SO_4 (V_1 ) " " = xml `
Normality of ` H_2SO_4 (N_1 )" " 1.5N `
Volume of `Ba(OH)_2(V_2) " " 35.8ml `
Normality of ` Ba(OH_2) (N_2) " " 2.5N`
According to volumetric law
` " " V_1 N_1 =V_2N_2`
` " " x xx1.5 = 35.8xx 2.5`
` " " x =( 35.8 xx 2.5)/( 1.5) `
` " " x= " " = 59.66ml `