Question

A.

B.

When 12.0 mL of a 5.46x10-4 M potassium carbonate solution is combined with 12.0 mL of a 7.70*10-4 M copper(II) nitrate solution does a precipitate form? (yes or no) For these conditions the Reaction Quotient, Q, is equal to

C.

Answer 1

A

To solve this problem, we must compare the Ksp value with Q value. So, we must need Ksp value for CuCoz. Ksp for Cuco, = 2.5 x 10-10 Number of moles of Cu+2 = M*V = 7.70 x 10-4 M * 12.0 mL = 0.00924 mmol Number of moles of CO2-2= M*V = 5.46 x 10-4 M * 12.0 mL = 0.006552 mmol Total volume = 12.0 + 12.0 = 24.0 mL After mixing, conc. of Cu+2 = [Cu+2] = moles/total volume = 0.00924 mmol/ 24.0 mL = 3.85 x 10-4 M. conc. of C03-2 = [CO3-2 ] = moles/total volume = 0.006552 mmol/ 24.0 mL = 2.73 x 10-4 M Q = [Cu+2][CO-2] = = 3.85 x 10-4 M x 2.73 x 10-4 M = 1.05 x 10-7 Ksp for CuCoz = 2.5 x 10-10 The Q is more than Ksp solubility product of CuCO3, Hence precipitation occurs. Answer: YES Q = 1.05 x 10-7

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B

To solve this problem, we must compare the Ksp value with Q value. So, we must need Ksp value for CuCoz. Ksp for Cuco, = 2.5 x 10-10 Number of moles of Cu+2 = M*V = 7.26 x 10-4 M * 12.0 mL = 0.008712 mmol Number of moles of CO2-2= M*V = 2.76 x 10-4 M * 22.0 mL = 0.006072 mmol Total volume = 12.0 + 22.0 = 34.0 mL After mixing, conc. of Cu+2 = [Cu+2] = moles/total volume = 0.008712 mmol/ 34.0 mL = 2.56 x 10-4 M. conc. of C03-2 = [CO3-2 ] = moles/total volume = 0.006072 mmol/ 34.0 mL = 1.79 ~ 10-4 M Q = [Cu+2][CO-2] = = 2.56 x 10-4 M x 1.79 x 10-4 M = 4.58 x 10-8 Ksp for CuCoz = 2.5 x 10-10 The Q is more than Ksp solubility product of CuCO3, Hence precipitation occurs. Answer: YES Q = 4.58 x 10-8

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C

To solve this problem, we must compare the Ksp value with Q value. So, we must need Ksp value for CaCO3. Ksp for CaCO3 = 3.8 x 10-9 Ken.. Number of moles of Ca+2 = M*V = 9.07 x 10-4 M * 18.0 mL = 0.016326 mmol Number of moles of CO2-2= M*V = 4.51 x 10-4 M * 12.0 mL = 0.005412 mmol Total volume = 18.0 + 12.0 = 30.0 mL After mixing, conc. of Ca+2 = [Ca+2] = moles/total volume = 0.016326 mmol/ 30.0 mL = 5.44 x 10-4 M. conc. of CO2-2 = [CO2-2] = moles/total volume = 0.005412 mmol/ 30.0 mL = 1.80 x 10-4 M Q = (Ca+21[C02-2] = = 5.44 x 10-4 M * 1.80 x 10-4 M = 9.82 x 10-8 Ksp for CaCO3 = 3.8 x 10-9 The Q is more than Ksp solubility product of CaCO3, Hence precipitation occurs. Answer: YES Q=9.82 x 10-8