Question 72818

Apr 8, 2017

K_text(sp) = 7.4 × 10^"-17"

Explanation:

You have a concentration cell:

$\text{Ag(s)"∣"Ag"^"+" (xcolor(white)(l) "mol/L")"∣∣""Ag"^"+"("0.135 mol/L")∣∣"Ag(s)}$

The reactions are:

Anode: $\textcolor{w h i t e}{m} \text{Ag(s)" → "Ag"^"+"(xcolor(white)(l) "mol/L") + e^"-}$
Cathode: $\text{Ag"^"+"("0.135 mol/L") + e^"-" → "Ag(s)}$
Cell: $\textcolor{w h i t e}{m m l} \text{Ag(s)" + "Ag"^"+"("0.135 mol/L") → "Ag"^"+"(xcolor(white)(l) "mol/L") "Ag(s)}$

E_text(cell) = E_text(cell)^° - (RT)/(nF)lnQ = 0 - (RT)/(nF)lnQ = -(RT)/(nF)lnQ = "0.180 V"

-(8.314 color(red)(cancel(color(black)("J·K"^"-1""mol"^"-1"))) × 298 color(red)(cancel(color(black)("K"))))/(1 × "96 485" color(red)(cancel(color(black)("J·V"^"-1"))))lnQ = 0.180 color(red)(cancel(color(black)("V")))

$\ln Q = \text{-7.010}$

$Q = {e}^{\text{-7.010" = 9.030 × 10^"-4}}$

Q =( ["Ag"^"+"]("products"))/( ["Ag"^"+"]("reactants")) = x/0.135 = 9.030 × 10^"-4"

x = 0.135 × 9.030 × 10^"-4" = 1.219 × 10^"-4"

∴ In the anode half-cell,

["Ag"^"+"] = x color(white)(l)"mol/L" = 1.219 × 10^"-4"color(white)(l) "mol/L"

["PO"_4^"3-"] = ⅓["Ag"^"+"] "mol/L" = ⅓xcolor(white)(l) "mol/L"

$\text{Ag"_3"PO"_4"(s)" ⇌ "3Ag"^"+""(aq)" + "PO"_4^"3-""(aq)}$

${K}_{\textrm{s p}} = {\left[\text{Ag"^"+"]^3["PO"_4^"3-"] = x^3 × ⅓x = ⅓x^4 = ⅓ × (1.219 × 10^"-4}\right)}^{4}$

= 7.4 × 10^"-17"#