Will precipitation occur when you add 0.05 mL of 0.10 M KBr to a saturated solution of AgCl?

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Will precipitation occur when you add 0.05 mL of 0.10 M KBr to a saturated solution of AgCl?

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Answer 1 · 2016-10-22T15:11:33

Yes, precipitation will occur.

Explanation:

We must first calculate the concentration of ${Ag}^{+}$ $"Ag"^+$ in the saturated solution of $AgCl$ $"AgCl"$ .

$AgCl(s) ⇌ {Ag}^{+} (aq) + {Cl}^{-} (aq); K_{sp} = 1.8 \times 10^{-10}$ $"AgCl(s)" ⇌ "Ag"^+"(aq)" + "Cl"^"-""(aq)"; K_text(sp) = 1.8 × 10^"-10"$
$m m m m m m m x m m m m m x$ $color(white)(mmmmmmm)x color(white)(mmmmm)x$

$K_{sp} = [{Ag}^{+}] [{Cl}^{-}] = x^{2} = 1.8 \times 10^{-10}$ $K_text(sp) = ["Ag"^+]["Cl"^"-"] = x^2 = 1.8 × 10^"-10"$

$x = 1.34 \times 10^{-5}$ $x = 1.34 × 10^"-5"$

∴ $[{Ag}^{+}] = 1.34 \times 10^{-5} l mol/L$ $["Ag"^+] = 1.34 × 10^"-5"color(white)(l) "mol/L"$

Add 0.05 mL of 0.10 mol/L $KBr$ $"KBr"$ to 1.0 mL of the saturated $AgCl$ $"AgCl"$ solution. Will a precipitate form?

$["Br"^"-"] = "0.10 mol/L" × (0.05 color(red)(cancel(color(black)("mL"))))/(1.05 color(red)(cancel(color(black)("mL")))) = 4.8 × 10^"-3" color(white)(l)"mol/L"$

$"AgBr(s)" ⇌ "Ag"^+"(aq)" color(white)(m)+color(white)(m)"Br"^"-""(aq)"; color(white)(m)K_text(sp) = 5.0 × 10^"-13"$
$color(white)(mmmmmll) 1.34 × 10^"-5" color(white)(mml)4.8 × 10^"-3"$

$Q_text(sp) = ["Ag"^+]["Br"^"-"] = 1.34 × 10^"-5" × 4.8 × 10^"-3" = 6.4 × 10^"-8"$

$Q_"sp" > K_"sp"$

∴ A precipitate will form.

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Will precipitation occur when you add 0.05 mL of 0.10 M KBr to a saturated solution of AgCl?

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