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

Oct 22, 2016

Yes, precipitation will occur.

#### Explanation:

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

$\text{AgCl(s)" ⇌ "Ag"^+"(aq)" + "Cl"^"-""(aq)"; K_text(sp) = 1.8 × 10^"-10}$
$\textcolor{w h i t e}{m m m m m m m} x \textcolor{w h i t e}{m m m m m} x$

K_text(sp) = ["Ag"^+]["Cl"^"-"] = x^2 = 1.8 × 10^"-10"

x = 1.34 × 10^"-5"

["Ag"^+] = 1.34 × 10^"-5"color(white)(l) "mol/L"

Add 0.05 mL of 0.10 mol/L $\text{KBr}$ to 1.0 mL of the saturated $\text{AgCl}$ 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"

$\text{AgBr(s)" ⇌ "Ag"^+"(aq)" color(white)(m)+color(white)(m)"Br"^"-""(aq)"; color(white)(m)K_text(sp) = 5.0 × 10^"-13}$
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"

${Q}_{\text{sp" > K_"sp}}$

∴ A precipitate will form.