# Silver Phosphate and Acetic Acid

Silver phosphate has the formula:

$Ag_{3}PO_{4}$

Acetic Acid has the formula

$CH_{3}COOH$

Mix them together and you get the equation:

$Ag_{3}PO_{4(s)} + 3\; CH_{3}COOH_{(aq)} \rightleftharpoons 3\; AgCH_{3}COO_{(s)} + H_{3}PO_{4(aq)}$

The equilibrium equation from this formula is:

$\kappa = \frac{[H_{3}PO_{4}]}{[CH_{3}COOH]^3}$

To compute the equilibrium constant from this, multiply by values that are equal to one, but that allow you to use known constants:

$\kappa = \frac{[H_{3}PO_{4}]}{[CH_{3}COOH]^3} * \frac{[H_{3}O^{+}]^3}{[H_{3}O^{+}]^3} * \frac{[CH_{3}COO^{-}]^3}{[CH_{3}COO^{-}]^3} * \frac{[Ag^{+}]^3}{[Ag^{+}]^3} * \frac{[HPO_{4}^{2-}]}{[HPO_{4}^{2-}]} * \frac{[PO_{4}^{3-}]}{[PO_{4}^{3-}]}$

$\kappa = \frac{(\kappa _{a(CH_{3}COOH)})^3 * \kappa_{sp(Ag_{3}PO_{4})}}{(\kappa _{sp(AgCH_{3}COO)})^3 * \kappa _{a(H_{2}PO_{4} ^{-})} * \kappa _{a(HPO_{4}^{2-})}}$

$\kappa = \frac{(1.8\cdot 10^{-5})^{3}(1.3\cdot 10^{-20})}{(2.0\cdot 10^{-3})^{3}(6.3\cdot 10^{-8})(4.2\cdot 10^{-13})}$

$\kappa = \frac{[H_{2}PO_{4}^{-}]}{[CH_{3}COOH]^2} =5.0\cdot 10^{-5}$

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### One response to “Silver Phosphate and Acetic Acid”

1. templarhawk

Thank you soo much for this! I love you soo much for helping me out