2 I am suprised that silver nitrate is this poorly soluble, compared to the nitrate -- and found this worth to add here. edited May 26 '17 at 16:45 Buttonwood 11k11 gold badge2222 silver badges4848 bronze badges Your choices are restrained as the precipitation of $$\ce{SO4^{2-}}$$ in $$\ce{BaSO4}$$ is the classical way to quantify the former and an electrochemical determination (in aqueous solution) is not practical. Electing $$\ce{Ba(OH)2}$$ may lead to the formation of silver hydroxyde, equally poorly soluble in water. $$\ce{Ba(PO3)2}$$ itself is very poorly soluble, as were the $$\ce{Ag3PO4}$$, too. As correctly stated by you, an aqueous solution of $$\ce{BaCl2}$$ is not suitable, as $$\ce{Ag+}$$ will form the precipitate of $$\ce{AgCl}$$. Hence I suggest to give $$\ce{Ba(NO3)2}$$ a try. The anion is the same as in $$\ce{AgNO3}$$. The solubility of this salt in water is reported to equal $$\pu{4.95 g / 100 mL }$$ ($$\pu{0 ^\circ{}C}$$); or even $$\pu{10.5 g/ 100 mL}$$ ($$\pu{25 ^\circ{}C}$$), respectively, according to the English entry in wikipedia. Addition: Please note, the solubility of $$\ce{Ag2SO4}$$ ($$\pu{0.83 g / 100 mL}$$ water) at $$\pu{25 ^\circ{}C}$$ (ref.) is substantially lower than the one of $$\ce{AgNO3}$$ ($$\pu{256 g / 100 mL}$$ water) at $$\pu{25 ^\circ{}C}$$ (ref). Your choices are restrained as the precipitation of $$\ce{SO4^{2-}}$$ in $$\ce{BaSO4}$$ is the classical way to quantify the former and an electrochemical determination (in aqueous solution) is not practical. Electing $$\ce{Ba(OH)2}$$ may lead to the formation of silver hydroxyde, equally poorly soluble in water. $$\ce{Ba(PO3)2}$$ itself is very poorly soluble, as were the $$\ce{Ag3PO4}$$, too. As correctly stated by you, an aqueous solution of $$\ce{BaCl2}$$ is not suitable, as $$\ce{Ag+}$$ will form the precipitate of $$\ce{AgCl}$$. Hence I suggest to give $$\ce{Ba(NO3)2}$$ a try. The anion is the same as in $$\ce{AgNO3}$$. The solubility of this salt in water is reported to equal $$\pu{4.95 g / 100 mL }$$ ($$\pu{0 ^\circ{}C}$$); or even $$\pu{10.5 g/ 100 mL}$$ ($$\pu{25 ^\circ{}C}$$), respectively, according to the English entry in wikipedia. Your choices are restrained as the precipitation of $$\ce{SO4^{2-}}$$ in $$\ce{BaSO4}$$ is the classical way to quantify the former and an electrochemical determination (in aqueous solution) is not practical. Electing $$\ce{Ba(OH)2}$$ may lead to the formation of silver hydroxyde, equally poorly soluble in water. $$\ce{Ba(PO3)2}$$ itself is very poorly soluble, as were the $$\ce{Ag3PO4}$$, too. As correctly stated by you, an aqueous solution of $$\ce{BaCl2}$$ is not suitable, as $$\ce{Ag+}$$ will form the precipitate of $$\ce{AgCl}$$. Hence I suggest to give $$\ce{Ba(NO3)2}$$ a try. The anion is the same as in $$\ce{AgNO3}$$. The solubility of this salt in water is reported to equal $$\pu{4.95 g / 100 mL }$$ ($$\pu{0 ^\circ{}C}$$); or even $$\pu{10.5 g/ 100 mL}$$ ($$\pu{25 ^\circ{}C}$$), respectively, according to the English entry in wikipedia. Addition: Please note, the solubility of $$\ce{Ag2SO4}$$ ($$\pu{0.83 g / 100 mL}$$ water) at $$\pu{25 ^\circ{}C}$$ (ref.) is substantially lower than the one of $$\ce{AgNO3}$$ ($$\pu{256 g / 100 mL}$$ water) at $$\pu{25 ^\circ{}C}$$ (ref). 1 answered May 26 '17 at 16:29 Buttonwood 11k11 gold badge2222 silver badges4848 bronze badges Your choices are restrained as the precipitation of $$\ce{SO4^{2-}}$$ in $$\ce{BaSO4}$$ is the classical way to quantify the former and an electrochemical determination (in aqueous solution) is not practical. Electing $$\ce{Ba(OH)2}$$ may lead to the formation of silver hydroxyde, equally poorly soluble in water. $$\ce{Ba(PO3)2}$$ itself is very poorly soluble, as were the $$\ce{Ag3PO4}$$, too. As correctly stated by you, an aqueous solution of $$\ce{BaCl2}$$ is not suitable, as $$\ce{Ag+}$$ will form the precipitate of $$\ce{AgCl}$$. Hence I suggest to give $$\ce{Ba(NO3)2}$$ a try. The anion is the same as in $$\ce{AgNO3}$$. The solubility of this salt in water is reported to equal $$\pu{4.95 g / 100 mL }$$ ($$\pu{0 ^\circ{}C}$$); or even $$\pu{10.5 g/ 100 mL}$$ ($$\pu{25 ^\circ{}C}$$), respectively, according to the English entry in wikipedia.