Thank you for reporting, we will resolve it shortly
Q.
$20 \,ml$ of $0.2 \, M \,Al _{2}\left( SO _{4}\right)_{3}$ is mixed with $20 \, ml$ of $6.6 \,M \,BaCl _{2}$, the concentration of $Cl^-$ ion in solution is
Solutions
Solution:
$2 \times 6.6 \times 20=M_{2} \times[20+20] \rightarrow$ Mixed with $20 \,mL$ of $Al _{2}\left( SO _{4}\right)_{3}$
It increases volume but can't effect concentration of $Cl ^{-}$because it does not have $Cl^-$
$M _{2}=\frac{2 \times 6.6 \times 20}{40}=6.6$
