2 I am suprised that silver nitrate is this poorly soluble, compared to the nitrate -- and found this worth to add here.
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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
source | link

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.