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.
