An aqueous solution contains an unknown concentration of Ba^(+). When 50 mL of 1 M solution of Na_(2)SO_(4) is added, BaSO_(4) is 1xx10^(-10). What is the original concentration of Ba^(2+)?

An aqueous solution contains an unknown concentration of Ba^(+). When

1.	An aqueous solution contains an unknown concentration of Ba^(+). When 50 mL of 1 M solution of Na_(2)SO_(4) is added, BaSO_(4) is 1xx10^(-10). What is the original concentration of Ba^(2+)?
Answer» `5xx10^(-9)M` `2xx10^(-9)M` `1.1xx10^(-9)M` `1.0xx10^(M)` Solution :As 50 mL of 1 M `Na_(2)SO_(4)` solution are present in 500 mL of final solution, this means that 1 M `Na_(2)SO_(4)` has been diluted 10 TIMES. Hence, Conc. Of `[SO_(4)^(2-)]` in final solution `= (1)/(10) M = 0.1 M` `K_(sp) ` of `BaSO_(4) = 10^(-10)` `:. [BA^(2+)][SO_(4)^(2-)]=10^(-10)` `[Ba^(2+)][0.1]=10^(-10)` or `[Ba^(2+)]=10^(-9)M` i.e., conc. of `Ba^(2+)` ions in final solution `=10^(-9)`M Volume of final solution = 500 mL Volume of ORIGINAL solution = 500 - 50 = 490 mL Applying `underset("(INITIAL)") M_(1)V_(1)=underset("(Final)")M_(2)V_(2)` `M_(1)xx900=(10^(-9))(500)` or, `M_(1)=1.11xx10^(-9)M`