Question

A piece of Al weighing $2.7$ g is titrated with $75.0$ mL of $H_{2}S{O}_4$ (specific gravity $1.18$ g $m{L}^{-1}$ and $24.7$% $H_{2}S{O}_4$ by weight). After the metal is completely dissolved, the solution is diluted to $400$ mL. Calculate the molarity of free $H_{2}S{O}_4$ solution.(give answer in the form of 10X)

Answer 1

mEq of $Al=$ mEq of $H_{2}S{O}_4$ reacted

mEq of $Al=\frac{2.7}{9}\times 1000=300$

(Eq.wt of $Al=\frac{Mw}{3}=\frac{27}{3}=9g)$

$(Al\to A{l}^{3+}+3e^{\ominus })$

Normality of $H_{2}S{O}_4=\frac{\text{percentage by weight}\times 10\times d}{E{w}_2}=\frac{24.7\times 10\times 1.18}{98/2}$ $=5.95$

mEq of $H_{2}S{O}_4=N\times$ Volume in mL $=5.95\times 75=446.25$

meq of Al added $=300$

meq of $H_{2}S{O}_4$ left after reaction $=446.25-300=146.25$

Solution is diluted to $400$ mL

$\therefore N_{H_{2}S{O}_4}left=\frac{meq}{mL}=\frac{146.25}{400}=0.0367$

$M_{H_{2}S{O}_4}left=\frac{N}{n}=\frac{0.0367}{2}=0.183$

Thus, the molarity of free sulphuric acid solution is $0.183$ M.