To `10 mL` of `1 M BaCl_(2)` solution `5mL` of `0.5 M K_(2)SO_(4)` is added. `BaSO_(4)` is precipitated out. What will happen?

To 10 mL of 1m BaCl_(2) solution 5mL of 0.5 M K_(2)SO_(4) is added BaSO_(4) is precipitated out. What will haapen?

10 mL of 1 M BaCl_(2) solution & 5 mL 0.5 M K_(2)SO_(4) are mixed together to precipitate out BaSO_(4) . The amount of BaSO_(4) precipated will be

10 mL of 1 M BaCl_(2) solution and 5 mL 0.5 N K_(2)SO_(4) are mixed together ot precipitate out BaSO_(4) . The amount of BaSO_(4) precipated will be

An aqueous solution contains an unknown concentration of Ba^(2+) . When 50 mL of 1 M solution of Na_(2)SO_(4) is added , BaSO_(4) just begins to precipitate . The final volume is 500 mL . The solubility product of BaSO_(4) is 1 xx 10^(-10) . what is the original concentration of Ba^(2+) ?

An aqueous solution contains an unknown concentration of Ba^(2+) . When 50 mL of a 1 M solution of Na_(2)SO_(4) is added, BaSO_(4 ) just begins to precipitate. The final volume is 500 mL. The solubility product of BaSO_(4) is 1xx10^(–10) . What is the original concentration of Ba^(2+) ?

An aqueous solution contains an unknown concentration of Ba^(2+) . When 50 mL of a 1 M solution of Na_(2)SO_(4) is added, BaSO_(4 ) just begins to precipitate. The final volume is 500 mL. The solubility product of BaSO_(4) is 1xx10^(–10) . What is the original concentration of Ba^(2+) ?