50 mL of $20.8 \%(w / V) \mathrm{BaCl}_2(a q)$ and 100 mL of $9.8 \%(w / V) \mathrm{H}_2 \mathrm{SO}

Question

50 mL of $20.8 \%(w / V) \mathrm{BaCl}_2(a q)$ and 100 mL of $9.8 \%(w / V) \mathrm{H}_2 \mathrm{SO}_4(a q)$ solutions are mixed. Molarity of $\mathrm{Cl}^{-}$ions in the resulting solution is: (At. mass of $\mathrm{Ba}=137$ )

Professor PC Thomas Classes · Answer

(b) $50 \mathrm{~mL} \mathrm{BaCl}_2(a q)$ solution contain 10.4 g $\mathrm{BaCl}_2$ $$ 	herefore \quad n_{\mathrm{BaCl}_2}=\frac{10.4}{137+71}=0.05 $$ moles of $\mathrm{H}_2 \mathrm{SO}_4$ in $100 \mathrm{~mL} \mathrm{H}_2 \mathrm{SO}_4$ (aq) solution $=\frac{9.8}{98}=0.1$ $\mathrm{BaCl}_2+\mathrm{H}_2 \mathrm{SO}_4 \longrightarrow \mathrm{BaSO}_4+2 \mathrm{HCl}$; moles of HCl formed $=0.1$ $$ \left[\mathrm{Cl}^{-}ight]=\frac{0.1 	imes 1000}{50+100}=0.666 \mathrm{M} $$