PbCl_2(s) is not very soluble in water. PbCl_2(s) Pb^2 + (aq) + 2 Cl^-(aq) If x

Question

PbCl_2(s) is not very soluble in water. PbCl_2(s) Pb^2 + (aq) + 2 Cl^-(aq) If x moles of PbCl_2(s) dissolve in 1.00 L of water, how many moles of Pb^2+(aq) are produced? How many moles of Cl^-(aq) are produced? The equilibrium constant, K_sp, for the dissolution of PbCl_2(s) in water is 1.6 Times 10^-5. What is the concentration of Pb^2+(aq) at equilibrium? What is the concentration of Cl^-(aq) at equilibrium? For each of the following situations, determine whether or not a precipitate of MgF_2 is expected to form. 500.0 mL of 0.050M Mg(N0_3)_2 is mixed with 500.0 mL of 0.010 M NaF. 500.0 mL of 0.050M Mg(NI_3)_2 is mixed with 500.0 mL of 0.0010 M NaF. Is a precipitate of Cd(CN)_2 expected to form when 500.0 mL of 0.010 M Cd(NO_3)_2 is mixed with 500.0 mL of 0.0025 M NaCN? Both cadmium(n) nitrate and sodium(l) cyanide are completely dissociated in the original solutions. The K_sp of Cd(CN)_2 is 1.0 Times 10^-8. The K_sp of Ag_2SO_4 is 1.4 Times 10^-5. Will a precipitate form when 250 mL of 0.12 M AgNQ_3 is mixed with 500 mL of 0.0050 M Na2SO_4? Write a chemical equation that describes the dissolution of solid AuCl_3 to Au^3+ and Cl^- ions found in water, Write the expression for the K_sp of AuCI_3 Calculate how many grams of Au^3+ would be found in one liter of a saturated solution of AuCl_3. The K_sp for AuCl_3 is 3.2 Times 10^-23.

Explanation / Answer

6. If x mole of PbCl2 in dissolved in 1 L of water

a) moles Pb2+ = x

moles of Cl- = 2x

b) Ksp = [Pb2+]]Cl-]^2

1.6 x10^-5 = (x)(2x)^2

x = 0.016 M

Pb2+ = 0.016 M

Cl- = 0.032 M

7. a) moles of Mg(NO3)2 = 0.05 x 0.5 = 0.025 mols

molarity = 0.025/1 = 0.025 M

moles of NaF = 0.01 x 0.5 = 0.005 mols

molarity = 0.005/1 = 0.005 M

So we will have 0.005 mols of prepipitate formed

Ksp = 5.6 x 10^-11

Q = [Mg2+][F-]^2

    = (0.025)(0.005)^2

    = 6.25 x 10^-7

Since Q is greater then Ksp preipitate will form.

b) moles of Mg(NO3)2 = 0.025 mols

molarity = 0.025 M

moles of NaF = 0.001 x 0.5 = 5 x 10^-4 M

Q = (0.025)(5 x 10^-4)^2

    = 6.25 x 10^-9

Since Q is greater then Ksp precipitate will form

8. Q = [Ca2+][CN]^2

moles of Cd(NO3)2 = 0.01 x 0.5 = 5 x 10^-3 mols

molarity of Cd(NO3)2 = 5 x 10^-3/1 = 5 x 10^-3 M

moles of NaCN = 0.01 x 0.5 = 5 x 10^-3 mols

molarity of NaCN = 5 x 10^-3 M

Q = (5 x 10^-3)(5 x 10^-3)^2

    = 1.25 x 10^-7

Q is greater than Ksp and thus precipitate would form.

Problems - Solutions

1. moles of AgNO3 = 0.12 x 0.250 = 0.03 mols

molarity of AgNO3 = 0.03/(0.25 + 0.5) = 0.04 M

moles of Na2SO4 = 0.005 x 0.5 = 2.5 x 10^-3 mols

molarity of Na2SO4 = 3.33 x 10^-3 M

Q = (0.08)^2(3.33 x 10^-3)

    = 2.13 x 10^-5

Q is greater than Ksp and thus precipitate would form

2. a) Chemical equation : AuCl3 <==> Au^3+ + 3Cl-

b) Ksp = [Au+][Cl-]^3

c) 3.2 x 10^-23 = (x)(3x)^3

x = 1.04 x 10^-6 M

moles of Au+ = 1.04 x 10^-6 mols

grams of Au3+ = 1.04 x 10^-6 x 196.96657 = 2.05 x 10^-4 g

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