What is the maximum mass (in grams) of each of the following soluble salts that can be added to 120 mL of 0.050 M BaCl2 without causing a precipitate to form? (Assume that the addition of solid causes no change in volume.)
(a) (NH4)2SO4
× 10
g
(Enter your answer in scientific notation.)
(b) Pb(NO3)2
g
(c) NaF
g
Here is how you do the first one. I'll be glad to check your work for the others. You can add enough of each salt so that the Ksp of the ppt is not exceeded.
(NH4)2SO4 + BaCl2 ==> BaSO4(s) + 2NH4Cl
The ppt will be BaSO4. Look up the Ksp for BaSO4. It is approx 1.1E-10 but you should use the value shown in your class/text book.
The Ksp expression is (Ba^2+)(SO4^2-) = 1.1E-10
The problem tells you that the BaCl2 is 0.05M. Plug that into the Ksp expression for (Ba^2+) and solve for (SO4)^2-. That will tell you the sulfate value that will just meet the Ksp for BaSO4. Anothr moleculer over that and the BaSO4 will ppt.
(0.05)(SO4^2-) = 1.1E-10
(SO4^2-) = 1.1E-10/0.05 = 2.2E-9 M = 2.2E-9 mols/L solution.
In 120 mL of solution there will be 2.2E-9 mols x (120/1000) = 2.64E-10 mols/120 mL. Convert that to grams (NH4)2SO4.
2.64E-10 mols x molar mass (NH4)2SO4 = grams (NH4)2SO4 in 120 mL =?
Good luck.
(a) (NH4)2SO4: Well, let me calculate that if there is a "precipitate" forming, then it must be a pretty fancy one. So, with my trusty calculator, I can tell you that the maximum mass of (NH4)2SO4 that can be added to 120 mL of 0.050 M BaCl2 without causing a precipitate to form is 5.38 x 10 grams. That's one stylish precipitate right there!
(b) Pb(NO3)2: Ah, lead nitrate, a classic choice! Let's see, the maximum mass of Pb(NO3)2 that can be added to 120 mL of 0.050 M BaCl2 without causing a precipitate to form... wait for it... is 8.55 grams. That's enough to get any clown's attention!
(c) NaF: Ah, the famous sodium fluoride, also known as the dentist's best friend! Now, let's calculate the maximum mass of NaF that can be added to 120 mL of 0.050 M BaCl2 without causing a precipitate to form. Drumroll, please... It is a whopping 4.41 grams. That's enough to make your teeth sparkle (but don't eat it, my friend)!
Remember, these calculations are just for fun and shouldn't be used in any serious scientific endeavors. Always consult your textbook or consult a qualified professional for accurate information.
To determine the maximum mass of each soluble salt that can be added without causing a precipitate to form, we need to use the concept of solubility product constant (Ksp). The Ksp is a measure of the solubility of a compound in water and is used to determine if a precipitate will form when the compound is added to a solution.
(a) (NH4)2SO4:
The balanced equation for the dissociation of (NH4)2SO4 is:
(NH4)2SO4 ⟶ 2NH4+ + SO4^2-
From the equation, we can see that one mole of (NH4)2SO4 produces 2 moles of NH4+ ions and one mole of SO4^2- ions.
According to the question, the concentration of BaCl2 is 0.050 M, and the volume is 120 mL. Therefore, the number of moles of BaCl2 is:
0.050 mol/L * 0.120 L = 0.006 mol BaCl2
Since there is a 1:1 stoichiometric ratio between BaCl2 and (NH4)2SO4, the maximum number of moles of (NH4)2SO4 that can be added without causing a precipitate is also 0.006 mol.
The formula weight of (NH4)2SO4 is:
[(2 x 14.01 g/mol) + (1 x 1.01 g/mol)] x 2 + (1 x 32.06 g/mol) + (4 x 16.00 g/mol) = 132.14 g/mol
To calculate the maximum mass, we multiply the number of moles by the formula weight:
0.006 mol x 132.14 g/mol = 0.7928 g
Therefore, the maximum mass of (NH4)2SO4 that can be added without causing a precipitate is 7.928 g.
(b) Pb(NO3)2:
The balanced equation for the dissociation of Pb(NO3)2 is:
Pb(NO3)2 ⟶ Pb^2+ + 2NO3^-
From the equation, we can see that one mole of Pb(NO3)2 produces one mole of Pb^2+ ions and two moles of NO3^- ions.
Following the same steps as above, the number of moles of Pb(NO3)2 is also 0.006 mol.
The formula weight of Pb(NO3)2 is:
(207.2 g/mol) + (2 x [14.01 g/mol + (3 x 16.00 g/mol)]) = 331.2 g/mol
Calculating the maximum mass:
0.006 mol x 331.2 g/mol = 1.9872 g
Therefore, the maximum mass of Pb(NO3)2 that can be added without causing a precipitate is 1.9872 g.
(c) NaF:
The balanced equation for the dissociation of NaF is:
NaF ⟶ Na+ + F-
Following the same steps as above, the number of moles of NaF that can be added is 0.006 mol.
The formula weight of NaF is:
22.99 g/mol + 18.99 g/mol = 41.98 g/mol
Calculating the maximum mass:
0.006 mol x 41.98 g/mol = 0.2519 g
Therefore, the maximum mass of NaF that can be added without causing a precipitate is 0.2519 g.
To determine the maximum mass of each soluble salt that can be added to 120 mL of 0.050 M BaCl2 without causing a precipitate to form, you can use the concept of solubility product constant (Ksp). The Ksp is a measure of the maximum concentration of ions in a solution before a precipitate forms.
First, let's determine the balanced equation for each salt when dissolved in water:
(a) (NH4)2SO4 dissociates into 2 NH4+ ions and 1 SO4^2- ion. The balanced equation is:
(NH4)2SO4 -> 2 NH4+ + 1 SO4^2-
(b) Pb(NO3)2 dissociates into 1 Pb^2+ ion and 2 NO3- ions. The balanced equation is:
Pb(NO3)2 -> 1 Pb^2+ + 2 NO3-
(c) NaF dissociates into 1 Na+ ion and 1 F- ion. The balanced equation is:
NaF -> 1 Na+ + 1 F-
Next, we need to find the solubility product constant (Ksp) for each salt. The Ksp can be found in reference books or online databases.
(a) The Ksp for (NH4)2SO4 is 1.6 x 10^-7 mol^2/L^2.
(b) The Ksp for Pb(NO3)2 is 1.4 x 10^-4 mol^3/L^3.
(c) The Ksp for NaF is 4.0 x 10^-11 mol^2/L^2.
Next, we need to calculate the maximum concentration of the common ion (Cl-) in the solution. BaCl2 dissociates into 1 Ba^2+ ion and 2 Cl- ions. Since the concentration of BaCl2 is 0.050 M, the concentration of Cl- ions in the solution is (2 x 0.050) = 0.100 M.
Now we can use the Ksp and the concentration of Cl- ions to calculate the maximum mass of each salt that can be added without causing a precipitate.
(a) For (NH4)2SO4:
The balanced equation shows that for each mole of (NH4)2SO4, you get 2 moles of NH4+ ions. Therefore, the maximum concentration of NH4+ ions is 2 x 0.050 = 0.100 M.
To find the mass, we can use the equation:
Ksp = [NH4+]^2 x [SO4^2-]
1.6 x 10^-7 = (0.100)^2 x [SO4^2-]
[SO4^2-] = 1.6 x 10^-7 / (0.100)^2 = 1.6 x 10^-7 / 0.01 = 1.6 x 10^-5 mol/L
To find the mass, we multiply the molar concentration by the volume of the solution:
mass = (1.6 x 10^-5 mol/L) x (0.120 L) x (142.04 g/mol)
mass ≈ 9.18 x 10^-5 g
Therefore, the maximum mass of (NH4)2SO4 that can be added is approximately 9.18 x 10^-5 grams.
(b) For Pb(NO3)2:
To find the concentration of Pb^2+ ions, we divide the concentration of Cl- ions (0.100 M) by 2 (as there is only 1 Pb^2+ ion for 2 Cl- ions in the balanced equation).
Pb^2+ concentration = 0.100 M / 2 = 0.050 M
To find the mass, we can use the equation:
Ksp = [Pb^2+] x [NO3-]^2
1.4 x 10^-4 = (0.050 M) x ([NO3-]^2)
[NO3-] = sqrt(1.4 x 10^-4 / 0.050) = 6.32 x 10^-3 M
To find the mass, we multiply the molar concentration by the volume of the solution:
mass = (6.32 x 10^-3 mol/L) x (0.120 L) x (331.2 g/mol)
mass ≈ 2.50 x 10^-2 g
Therefore, the maximum mass of Pb(NO3)2 that can be added is approximately 2.50 x 10^-2 grams.
(c) For NaF:
To find the concentration of F- ions, we divide the concentration of Cl- ions (0.100 M) by 2 (as there is only 1 F- ion for 2 Cl- ions in the balanced equation).
F- concentration = 0.100 M / 2 = 0.050 M
To find the mass, we can use the equation:
Ksp = [Na+] x [F-]
4.0 x 10^-11 = (0.050 M) x [F-]
[F-] = 4.0 x 10^-11 / 0.050 = 8.0 x 10^-10 M
To find the mass, we multiply the molar concentration by the volume of the solution:
mass = (8.0 x 10^-10 mol/L) x (0.120 L) x (41.99 g/mol)
mass ≈ 3.19 x 10^-9 g
Therefore, the maximum mass of NaF that can be added is approximately 3.19 x 10^-9 grams.