100 g of HCI solution with relative density 1.117 g/ml contains 33.4 g of HCI. What volume of this HCI solution will be required to exactly neutralize 5 litre of N/10 NaOH solution?

Explanation

Volume of HCl solution ` = 100/(1.17) mL `
`(" density " = ("mass")/("volume"))`
Equivalents of HCl = `(33.4)/(36.5)`
(eq.wt . Of HCl = 36.5)
m.e of HCl ` = (33.4)/(36.5) xx 1000 " " … (Eqn . 3)`
Normality of HCl = `("m.e")/(" volume in mL") " " ....(Eqn .1)`
` =(33.4)/(36.5) xx 1000 xx (1.17)/100`
` = 10.7 ` N
Now let the volume of HCl, of normality calculated above , required to neutralise exactly the given NaOH solution be v mL .
Now let the volume of HCl , of normality calculated above , required to neutralise exactly the given ,naOH solution be v mL .
m.e of HCl = m.e of NaOH
`10.7 xx v = 1/10 xx 5000`
` 10 . 7 xx v 1/10 xx 5000`
`10.7 xx v = 500`
` :. " " v = 46.7 mL `