A 2.00L volume of a 0.0200M solution of AgNO3 is combined with 2.00L of 0.400M N

Question

A 2.00L volume of a 0.0200M solution of AgNO3 is combined with 2.00L of 0.400M Na2CrO4. Ksp = 1.1*10^-12 M^3. Find the following;

The answers are a. Q = 2.00*10^-5 M^3 b. NO3 = 0.01M Na = 0.400M CrO4 = 0.195M Ag = 2.5*10^-6M c. 2.5*10^-2 percent left

A 2.00 L volume of a 0.0200 M solution of AgNO3(aq) is combined with 2.00 L of 0.400 mol L Na2Cro4(aq). Ksp Ag2Cro4 1.1 x 10 12 mol L 3. (a) Show that Ag2Cro.(s) forms in this solution. (b) Calculate the [Ag [NO2 [Nat] and Cro4-1 in the combined solution after precipitation is complete (c) What of the Ag remains in solution after the precipitation of AgeCro is complete?

Explanation / Answer

Ag2CrO4 <== 2Ag+ + CrO4^2-

Ksp = [Ag+]^2[CrO4^2-]

concentration of Ag+ in the solution = 0.02 * 2L/4L = 0.01 M

Concentration of CrO4^2- = 0.4 M * 2L/4L = 0.2 M

Ionic product = [Ag+]^2[CrO4^2-] = (0.01 M)^2 * 0.2 M = 2*10^-5 M

As ionic product is greater than Ksp , Ag2CrO4 will be precipitated from the solution.

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Amount of NO3^- = 0.02 * 2L/ 4L = 0.01 M

Na2CrO4 <==> 2Na+ + CrO4^2-

amount of Na+ = (2 * 0.4 M ) * 2L/4L = 0.4 M

2AgNO3 + Na2CrO4 = Ag2CrO4 + NaNO3

moles of AgNO3 present = 0.02 M * 2L = 0.04 moles

moles of Na2CrO4 present = 0.4 * 2L = 0.8 moles

So, 2 moles AgNo3 reacts with 1 mol Na2CrO4. Amount of Na2CrO4 that will react with AgNO3 is 0.02 moles

Moles of excess CrO4^2- = 0.8 -0.02 = 0.78 moles

After the precipitation is complete, some of the Ag2CrO4 will redissolve in water .

                             Ag2CrO4 <==> 2Ag+ + CrO4^-

initial                           0                 0            0.78

change                                          +2x            +x

equilibrium                                     2x               0.78+x

Ksp =(2x)^2(0.76 + x)

or, 1.1*10^-12 = (2x)^2(0.78 + x)

As x is very small 0.78 +x = 0.78

1.1*10^-12 = (2x)^2(0.78)

or, x = 5.94*10^-7 moles

Concentration Ag+ afetr precipitation =2x = 1.18 *10^-6 moles

Molarity of Ag+ after precipitation = 1.18 *10^-6 moles/4L = 2.97*10^-7 M

Concentration of CrO4- after precipitation = Ksp /x^2 = 0.78 M

% remaining = moles of Ag+ after precipitation /moles of Ag+ before precipitation * 100%

                    = [2.97*10^-7 M* 4L /0.02 * 4L] * 100 = 1.45*10^-3 %

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