1. What mass of lead chloride (PbCl2, MW= 278.2; Ksp= 1.8 x 10^-10) is dissolved in 250 mL of a saturated solution?
2. What mass lead chloride (PbCl2, MW=278.2; Ksp=1.8 x 10^-10) will dissolve in 150 mL of 0.050 M Pb(NO3)2?
PbCl2 ==> pb^+2 + 2Cl^-
Ksp + (Pb^+2)(Cl^-)^2
Set up ICE chart with Pb^+2 = x and Cl^- = 2x, substitute into Ksp and solve for x. x will have units of moles/L; therefore, convert moles/L to grams (g = moles x molar mass) and divide by 4 to find the mass PbCl2 in 250 mL.
For #2, you have a common ion of Pb^+2 from Pb(NO3)2.
PbCl2 ==> Pb^+2 + 2Cl^-
Ksp = etc.
Here Pb^+2 comes from two sources; i.e., PbCl2 and Pb(NO3)2. I think you can ignore the Pb^+2 from PbCl2. Substitute M Pb from Pb(NO3)2, solve for Cl^-, convert to PbCl2 (by taking half of the Cl^-), convert to grams PbCl2, (that will be grams/L), then convert to g/150 mL.
Post your work if you get stuck.
Thank you so much for helping, but I'm confused to why I'm diving it by 4 on the first one?
Ksp=1.8 x 10^-10=(2x)^2(x)
1.8 x 10^-10=4x^3
Divide by 4= 4.5 x 10^-11
Cube Root= 3.556893304 x 10^-4 mols/L
Times Molar Mass: 0.0989527717
Divide by 4: 0.0247381929, but why?
Why divide by 4? Your answer of 0.09895 grams PbCl2 (the 3.55 x 10^-4 is M is in moles/L.) gives grams/liter of solution. The problem asks for the solubility of PbCl2 in grams/250 mL and 250 mL is 1/4 L.
By the way, you have carried out the number of decimal places far to many. The Ksp has two significant figures in it; therefore, you must round the answer you have to two places. I would round 0.02474 to 0.025 g/250 mL.
Does any solid pbcl2 form when 3.5 mg of NaCl is dissolved in 0.25L of 0.12M Pb(NO3 )2
1. Well, lead chloride sure sounds clingy. Let's see how much of it is dissolved in that 250 mL of liquid love. To find out, we need to calculate the concentration of PbCl2 in the saturated solution.
First, let's convert the volume to liters: 250 mL = 0.25 L.
Now, we can use the solubility product constant (Ksp) to calculate the concentration. The equation for the dissociation of PbCl2 is PbCl2 --> Pb2+ + 2Cl-. Since the stoichiometry is 1:1, the concentration of Pb2+ is the same as the concentration of PbCl2.
Ksp = [Pb2+][Cl-]^2
Since [Pb2+] = [PbCl2], we can rewrite the equation as: Ksp = [PbCl2][Cl-]^2
Now, let's solve for [PbCl2]. Since the concentration of Pb2+ is equal to the concentration of PbCl2, we'll call it x. The concentration of Cl- will be 2x because of the stoichiometric coefficient of 2 in the equation.
Ksp = x * (2x)^2
Simplifying this, we get: Ksp = 4x^3
Now, let's substitute the value of Ksp: 1.8 x 10^-10 = 4x^3
After solving for x, we find x = 3.77 x 10^-4 M. This is the concentration of PbCl2 in the saturated solution.
To find the mass of PbCl2, we multiply the concentration by the volume: mass = concentration * volume = 3.77 x 10^-4 M * 0.25 L = 9.42 x 10^-5 moles.
Finally, we can find the mass of PbCl2 by multiplying the number of moles by the molecular weight: mass = 9.42 x 10^-5 moles * 278.2 g/mol ≈ 0.0262 grams.
So, approximately 0.0262 grams of lead chloride is dissolved in that 250 mL of liquid love.
2. Well, well, well, it looks like you want to know how much lead chloride will flirt with that 150 mL of 0.050 M Pb(NO3)2. Let's crunch the numbers and calculate the mass, shall we?
First, we need to determine the moles of Pb(NO3)2 in the solution. We start by converting the volume to liters: 150 mL = 0.150 L.
The molarity (M) of Pb(NO3)2 is given as 0.050 M, so we can calculate the number of moles using the equation: moles = concentration * volume.
moles = 0.050 M * 0.150 L = 0.0075 moles.
Now, we can use the stoichiometry of the balanced equation to find the moles of PbCl2 that will be formed.
The equation for the reaction between Pb(NO3)2 and PbCl2 is: Pb(NO3)2 + 2Cl- --> PbCl2 + 2NO3-
Here, the stoichiometry is 1:1, so the moles of PbCl2 formed will be equal to the moles of Pb(NO3)2 used.
Therefore, the mass of PbCl2 that will dissolve is 0.0075 moles * 278.2 g/mol = 2.09 grams.
Approximately 2.09 grams of lead chloride will dissolve in that 150 mL of 0.050 M Pb(NO3)2 solution.
To find the mass of lead chloride (PbCl2) dissolved in a saturated solution, you need to use the solubility product constant (Ksp) and the volume of the solution.
1. To find the mass of PbCl2 dissolved in 250 mL of a saturated solution:
Step 1: Calculate the concentration of PbCl2 in the saturated solution.
- The Ksp for PbCl2 is given as 1.8 x 10^-10.
- Assume that x moles of PbCl2 dissolve in 1 liter of the saturated solution.
- So, the concentration of PbCl2 is x moles/L, or x M (molar concentration).
Step 2: Convert the volume of the solution from mL to L.
- 250 mL = 250/1000 L = 0.25 L
Step 3: Calculate the number of moles of PbCl2 in the solution using the molar concentration and volume.
- Moles of PbCl2 = concentration (M) x volume (L) = x M * 0.25 L = 0.25x moles
Step 4: Calculate the mass of PbCl2 using its molar mass (MW).
- Mass = moles * MW = 0.25x moles * 278.2 g/mol = 69.55x grams
Therefore, the mass of PbCl2 dissolved in 250 mL of a saturated solution is 69.55x grams.
2. To find the mass of PbCl2 that will dissolve in 150 mL of 0.050 M Pb(NO3)2:
Step 1: Calculate the number of moles of Pb(NO3)2 using its molar concentration and volume.
- Moles of Pb(NO3)2 = concentration (M) x volume (L) = 0.050 M * 0.150 L = 0.0075 moles
Step 2: Determine the stoichiometry of the reaction between Pb(NO3)2 and PbCl2. The balanced equation is:
Pb(NO3)2 + 2NaCl -> PbCl2 + 2NaNO3
Step 3: Use the stoichiometry ratio to find the number of moles of PbCl2 that will be formed.
- From the balanced equation, 1 mole of Pb(NO3)2 produces 1 mole of PbCl2.
- Therefore, 0.0075 moles of Pb(NO3)2 would produce 0.0075 moles of PbCl2.
Step 4: Calculate the mass of PbCl2 using its molar mass (MW).
- Mass = moles * MW = 0.0075 moles * 278.2 g/mol = 2.0825 grams
Therefore, the mass of PbCl2 that will dissolve in 150 mL of 0.050 M Pb(NO3)2 is 2.0825 grams.