We must first calculate the concentration of "Ag"^+ in the saturated solution of "AgCl".
"AgCl(s)" ⇌ "Ag"^+"(aq)" + "Cl"^"-""(aq)"; K_text(sp) = 1.8 × 10^"-10"
color(white)(mmmmmmm)x color(white)(mmmmm)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 "KBr" to 1.0 mL of the saturated "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.