the answer is not 5.4*10^8 To a 100.0 mL volumetric flask are added 1.00 mL volu

Question

the answer is not 5.4*10^8

To a 100.0 mL volumetric flask are added 1.00 mL volumes of three solutions: 0.0100 MAgNO3, 0.245 M NaBr, and 0.100 M NaCN. The mixture is diluted with deionized water to the mark and shaken vigorously. What mass of AgBr would precipitate from this mixture? (Hint: The K, of AgBr is 5.4 x 10^-13 and the Kf of Ag(CN)2 is 1.0 X 10^21) Analyze Using the dilution equation, we first calculate the concentrations of the reacting ions (Ag+, Br-, and CN-): The pertinent solubility and complex formation equations are:

Explanation / Answer

AgNO3   <----> Ag+  + NO3- [AgNO3] = [Ag+] = [NO3-]

C1 * V1 = C2 * V2

0.01 M * 0.001 L = C2 * 0.1 L ---> C2 = 0.0001 M = [Ag+] = [NO3-]

NaBr <-----> Na+   + Br - [NaBr] = [Na+] = [Br-]

C1 * V1 = C2 * V2

0.245 M * 0.001 L = C2 * 0.1 L ---> C2 = 0.00245 M = [Na+] = [Br-];

NaCN <----> Na+ + CN-

C1 * V1 = C2 * V2

0.1 M * 0.001 L = C2 * 0.1 L ---> C2 = 0.001 M = [Na+] = [CN-]

Now, there are two reactions here:

Ag+   + 2CN- ---> Ag(CN)2-

initial: 0.0001 0.001 -

change -x -2x +x

end 0.0001-x 0.001-2x x

Kf = x / (0.0001-x) * (0.001-2x)2 = 1*1021

x = 0.0000946

Ag+ left = 0.0001 - 0.0000946 = 0.0000054 M

Ag+  + Br -  <----> AgBr

Ksp = [Ag+] [Br-] ---> [Br-] = Ksp / [Ag+] = 1*10-7 M

So, you had 0.00245 M in the begining but you only need 1*10-7 M, so:

Br- leftover = 0.00245 - 1*10-7 = 0.0024499 M or 0.0024499 moles

AgBr Molar Mass= 187.77 g/mol

AgBr precipitate = 0.0024499 moles * 187.77 g/mol = 0.458 g

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