`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

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

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 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 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?

100 mL of 0.003 M BaCl_2 solution is mixed with 200 ml of 0.0006 M H_2SO_4 solution. Predict whether a precipitate of BaSO_4 will be formed or not. K_(sp) " for " BaSO_4 " is " 1.1 xx 10^(-10)

Give reason why BaSO_(4) will precipitate out when equal volumes of 2xx10^(-3) M BaCl_(2) solution and 2xx10^(-4)M Na_(2)SO_(4) solution are mixed. Given that the solubility product of BaSO_(4) is 1xx10^(-10).

Give reason why BaSO_(4) will precipitate out when equal volumes of 2xx10^(-3) M BaCl_(2) solution and 2xx10^(-4)M Na_(2)SO_(4) solution are mixed. Given that the solubility product of BaSO_(4) is 1xx10^(-10).