# Answer to Question #27035 in Other Chemistry for sam

Question #27035

will a precipitate occur if 250 ml of 1.5 x 10^.3 M NaCl is mixed with 250 mL of 2.0 x 10^.7 M AgNO3? Ksp of AgCl = 1.6 x 10^-10

Expert's answer

The equation for the chemical reaction is

NaCl(aq) + AgNO

K

From the equation for K

[Ag

The amount of NaCl is

n(NaCl) = C(NaCl) x V(L) = 1.5 x 10^-3 x 0.25 = 3.75 x 10^-4 mol

The amount of AgNO

n(AgNO

The precipitate will occur if the product of [Ag

[Ag

[Cl

[Ag

Answer: The precipitate won't occur.

NaCl(aq) + AgNO

_{3}(aq) = AgCl(s) + NaNO_{3}(aq)K

_{sp}= [Ag^{+}] x [Cl^{-}] = 1.6 x 10^-10From the equation for K

_{sp}:[Ag

^{+}] = [Cl^{-}] = (1.6 x 10^-10) ^0.5 = 1.3 x 10^-5 MThe amount of NaCl is

n(NaCl) = C(NaCl) x V(L) = 1.5 x 10^-3 x 0.25 = 3.75 x 10^-4 mol

The amount of AgNO

_{3}isn(AgNO

_{3}) = C(AgNO_{3}) x V(L) = 2.0 x 10^-7 x 0.25 = 5.0 x 10^-8 molThe precipitate will occur if the product of [Ag

^{+}] and [Cl^{-}] will be more or equal to K_{sp}:[Ag

^{+}] = n(Ag^{+}) / V(of new solution) = 5.0 x 10^-8 / (0.25 + 0.25) = 1.0 x 10^-7 M[Cl

^{-}] = n(Cl^{-})/V(of new solution) = 3.75 x 10^-4 / (0.25 + 0.25) = 7.5 x 10^-4 M[Ag

^{+}] x [Cl^{-}] = 1.0 x 10^-7 x 7.5 x 10^-4 = 7.5 x 10^-11 it is less than K_{sp}, that's why the precipitate will not occurAnswer: The precipitate won't occur.

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