A saturated solution of lead ii trioxonitrte v was prepared at 22degree Celsius at 27cm3 of this solution required 46cm3 of NaCl solution containing 96gdm3 for complete precipitation. Find the solubility of the lead in trioxonitrte v at 22degree Celsius in (a) mol/dm3 (b) g/dm3.
Pb(NO3)2 + 2NaCl ==> PbCl2 + 2NaNO
mols NaCl = grams/molar mass = 96/58.45 = approx 1.6 but you should confirm this as well as all other calculations that follow. I've estimated.
1.6 mols/dm3 = 1.6 M
mmols NaCl = mL x M = 46 x 1.6 = approx 75.
That means the amount of Pb(NO3)2 in solution is 1/2 that (look at the coefficients in the balanced equation) or 75/2 = approx 37.
mols Pb(NO3)2 = 0.037
27 cc = 0.027 cm^3
M Pb(NO3)2 = mols/dm^3 = ?
for part b convert. Im sure you can do that.
To find the solubility of lead (II) trioxonitrate (V) in mol/dm³ and g/dm³, we need to use the given information about the volumes and concentrations of the solutions involved.
Step 1: Calculate the number of moles of NaCl used
First, we need to find the number of moles of NaCl used. We can use the given concentration and volume of the NaCl solution.
The concentration of NaCl solution is 96 g/dm³, and the volume used is 46 cm³.
To convert the volume to dm³, divide by 1000: 46 cm³ ÷ 1000 = 0.046 dm³.
Now we can calculate the number of moles of NaCl:
Number of moles of NaCl = concentration × volume
= 96 g/dm³ × 0.046 dm³
= 4.416 g
Step 2: Calculate the number of moles of Pb(NO₃)₂
Since the reaction between Pb(NO₃)₂ and NaCl is 1:2, we can calculate the number of moles of Pb(NO₃)₂ using the moles of NaCl.
Number of moles of Pb(NO₃)₂ = (2/1) × number of moles of NaCl
= 2 × 4.416 mol
= 8.832 mol
Step 3: Calculate the volume of the saturated solution
We are given that 27 cm³ of the saturated solution is required to precipitate all the Pb(NO₃)₂.
Step 4: Calculate the solubility in mol/dm³
The solubility of Pb(NO₃)₂ can be calculated by dividing the number of moles by the volume in dm³.
Solubility of Pb(NO₃)₂ = number of moles / volume in dm³
= 8.832 mol / (27 cm³ ÷ 1000)
= 327.1 mol/dm³
Step 5: Calculate the solubility in g/dm³
To find the solubility of Pb(NO₃)₂ in g/dm³, we need to know the molar mass of Pb(NO₃)₂.
The molar mass of Pb(NO₃)₂ = (207.2 g/mol) + 2 × [(14.0 g/mol) + (3 × 16.0 g/mol)]
= 331.2 g/mol
Now we can calculate the solubility in g/dm³:
Solubility of Pb(NO₃)₂ = solubility in mol/dm³ × molar mass
= 327.1 mol/dm³ × 331.2 g/mol
= 108,247.52 g/dm³
Therefore, the solubility of lead (II) trioxonitrate (V) at 22°C is:
(a) 327.1 mol/dm³
(b) 108,247.52 g/dm³
To find the solubility of lead II trioxonitrate V at 22 degrees Celsius, we can use the equation:
Pb(NO3)2 (aq) + 2NaCl (aq) -> PbCl2 (s) + 2NaNO3 (aq)
In this equation, we can see that Pb(NO3)2 reacts with NaCl to form PbCl2, which is a precipitate. The volume of NaCl solution required for complete precipitation can be used to determine the solubility of Pb(NO3)2.
Given:
Volume of NaCl solution required = 46 cm3
Concentration of NaCl solution = 96 g/dm3
To find the solubility of Pb(NO3)2 at 22 degrees Celsius, we need to convert the given volume to dm3 and then find the amount of Pb(NO3)2 that was dissolved in that volume.
Step 1: Convert the volume of NaCl solution from cm3 to dm3.
Volume in dm3 = Volume in cm3 / 1000
Volume in dm3 = 46 cm3 / 1000
Volume in dm3 = 0.046 dm3
Step 2: Calculate the amount in moles of NaCl using the given concentration.
Amount of NaCl (in moles) = Concentration x Volume
Amount of NaCl (in moles) = 96 g/dm3 x 0.046 dm3
Amount of NaCl (in moles) = 4.416 moles
Step 3: Calculate the amount of Pb(NO3)2 in moles using the stoichiometry of the balanced equation.
From the balanced equation, we can see that 2 moles of NaCl react with 1 mole of Pb(NO3)2.
Therefore, Amount of Pb(NO3)2 (in moles) = (Amount of NaCl) / 2
Amount of Pb(NO3)2 (in moles) = 4.416 moles / 2
Amount of Pb(NO3)2 (in moles) = 2.208 moles
Now, we can find the solubility of Pb(NO3)2 at 22 degrees Celsius.
(a) Solubility in mol/dm3:
Solubility = Amount of Pb(NO3)2 (in moles) / Volume (in dm3)
Solubility = 2.208 moles / 0.027 dm3
Solubility = 81.78 mol/dm3
(b) Solubility in g/dm3:
To find the solubility in g/dm3, we need to consider the molar mass of Pb(NO3)2.
Molar mass of Pb(NO3)2 = (1 x Atomic mass of Pb) + (2 x Atomic mass of N) + (6 x Atomic mass of O)
Molar mass of Pb(NO3)2 = (1 x 207.2) + (2 x 14) + (6 x 16)
Molar mass of Pb(NO3)2 = 331.2 g/mol
Solubility (in g/dm3) = Solubility (in mol/dm3) x Molar mass of Pb(NO3)2
Solubility (in g/dm3) = 81.78 mol/dm3 x 331.2 g/mol
Solubility (in g/dm3) = 27,073.376 g/dm3
Therefore, the solubility of lead II trioxonitrate V at 22 degrees Celsius is:
(a) 81.78 mol/dm3
(b) 27,073.376 g/dm3