Question

what is [Ag^+] in a solution formed by mixing 25.0 ml of 0.10 M AgNO3 with 50.0 ml of 1.50 M Na3PO4. Ksp for Ag3PO4=1.0x10^-21

Answer 1

Answer on Question #56433 - Chemistry - Inorganic Chemistry

Question:

What is $\left[\mathrm{Ag}^{+}\right]$ in a solution formed by mixing $25.0~\mathrm{mL}$ of $0.10\mathrm{M}$ $\mathrm{AgNO}_3$ with $50.0~\mathrm{mL}$ of $1.50\mathrm{M}$ $\mathrm{Na}_3\mathrm{PO}_4$. $K_{\mathrm{sp}}$ for $\mathrm{Ag}_3\mathrm{PO}_4 = 1.0 \times 10^{-21}$?

Answer:

$$3 \mathrm{AgNO}_3 + \mathrm{Na}_3\mathrm{PO}_4 = \mathrm{Ag}_3\mathrm{PO}_4 + 3 \mathrm{NaNO}_3$$ $$3 \mathrm{Ag}^+ + \mathrm{PO}_4^{3-} = \mathrm{Ag}_3\mathrm{PO}_4$$

Volume of the solution after mixing is $25.0~\mathrm{mL} + 50.0~\mathrm{mL} = 75.0~\mathrm{mL}$;

$$n_{\mathrm{Ag}^{+}} = 0.10~\mathrm{M} * 25.0~\mathrm{mL} = 0.0025~\mathrm{mol};$$ $$n_{\mathrm{PO43}^{-}} = 1.50~\mathrm{M} * 50.0~\mathrm{mL} = 0.075~\mathrm{mol};$$

Each mole of $\mathrm{PO}_4^{3-}$ is precipitated by three moles of $\mathrm{Ag}^{+}$.

If quantity of $\mathrm{Ag}^{+}$ is $x\mathrm{mol}$, quantity of precipitated $\mathrm{Ag}^{+}$ is $(0.0025 - x)\mathrm{mol}$ and quantity of $\mathrm{PO}_4^{3-}$ in solution is $(0.075 - 3*(0.0025 - x)) = (0.0675 + 3x)\mathrm{mol}$

Concentrations are:

$$[\mathrm{Ag}^+] = x / 0.075$$ $$[\mathrm{PO}_4^{3-}] = (0.0675 + 3x) / 0.075$$

Product of solubility:

$$K_{\mathrm{sp}} = [\mathrm{Ag}^+]^{3}[\mathrm{PO}_4^{3-}] = 1.0 * 10^{-21} = (x / 0.075)^{3} \left( (0.0675 + 3x) / 0.075 \right)$$ $$x = 7.78 * 10^{-9}~\mathrm{mol} \approx 7.8 * 10^{-9}~\mathrm{mol}$$ $$[\mathrm{Ag}^+] = 1.04 * 10^{-7}~\mathrm{M} \approx 1.0 * 10^{-7}~\mathrm{M}$$

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