Consider the equilibrium AgCl(s) Ag + (aq) + Cl (aq) K c = 1.8x10 10 If 500 mL o

Question

Consider the equilibrium

                                    AgCl(s) Ag+(aq) + Cl(aq)             Kc = 1.8x1010

If 500 mL of 2x104 M NaCl and 500 mL of 2x105 M AgNO3 are mixed. How many moles of AgCl will precipitate? (Hint: don’t forget what happens to the concentrations when you mix the solutions.)

*          a)         8x106 moles ( this is the correct answer but i do not understand how it ws calculated)

            b)         2.0x105 moles

              c) 8x107 moles

            d)         2.0x104 moles

            e)         1.2x108 moles

Explanation / Answer

[ Ag+ ] = 2 ×10^-5/2 = 1× 10^-5M

[ Cl- ] = 2× 10^ - 4/2 = 1 × 10^ -4M

Ksp = 1.8×10^-10

Ksp = [ Ag+] [ Cl- ]

1.8 × 10^-10 = [ Ag+] × [ 1× 10^-4]

[ Ag+ ] = 1.8 × 10^-6

mole of Ag+ dissolved = 1.8 × 10^-6

Mole of Ag+ undissolved = (1×10^-5)-(1.8 × 10^-6)=8.2 × 10^-6

No of mol of AgCl formed = 8 × 10^-6

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