Calculate the mass of hydrochloric acid in `200 cm^(3)` of 0.2 N solution of it. What volume of this acid solution will react exactly with `25 cm^(3)` of 0.14 N solution of sodium hydroxide?

Explanation

Normality of hydrochloric acid = 0.2, Eq. mass of hydrochloric acid = 36.5
Mass of hydrochloric acid in one `dm^(3)` of the solution is given by
`W = N xx E = 0.2 xx 36.5` g.
`therefore` mass of hydrochloric acid in `200 cm^(3)` of the solution `=(0.2 xx 36.5)/5 = 1.46` g
`underset("Acid")(V_(1) xx N_(1)) = underset("base")(V_(2) xx N_(2))` Substituting we have `V_(1) xx 0.2 = 25 xx 0.14`
`therefore V_(1) = (25 xx 0.14)/0.2 = 17.5`
`therefore` Volume of the acid solution required `= 17.5 cm^(3)`.