Calculate the volume of 1.5 N ` H_2SO_4 ` is completely neturalized by 35.8 mL of
` 2.5 N NaOH`

Explanation

`V_1 N_1 =V_2N_2 `
Volume of ` H_2SO_4 (V_1) " "xml `
Normality of ` H_2SO_4 (V_1) " " 1.5N`
Volume of `NaOH (V_2) " "= 35.8mL`
Normalitiy of ` NaOH(N_2) " " 2.5N `
According to volumetric law
` " "V_1 N_1 =V_2N_2`
` x xx 1.5 = 35.8 xx 2.5 `
` " " x =( 35.8xx 2.5)/( 1.5) `
`" " x" " = 59.66mL `