The concentration of lead ions (Pb2+) in a sample of
polluted water that also contains nitrate ions
(NO3−) is determined by adding solid sodium
sulfate (Na2SO4) to exactly 500 mL of the
water. Calculate the molar concentration of Pb2+ if
0.00230 g of Na2SO4 was needed for the
complete precipitation of lead ions as PbSO4.
The reaction between Pb2+ and Na2SO4 is as follows
Pb2+(aq) + Na2SO4 (aq) ------> PbSO4(s) + 2Na+(aq)
stoichiometrically, 1mole of Na2SO4 react with 1mole of Pb2+
moles of Na2SO4 consumed = 0.00230g/142.04g/mol = 1.62×10-5
number of moles of Pb2+ = 1.62×10-5 mol
molarity of Pb2+ =(1.62 × 10-5mol/500ml)×1000ml
Molarity of Pb2+ = 3.24×10-5 M
The concentration of lead ions (Pb2+) in a sample of polluted water that also contains nitrate...
Enter your answer in the provided box. The concentration of lead ions (Pb2) in a sample of polluted water that also contains nitrate ions (NO3 ) is determined by adding solid sodium sulfate (Na2SO4) to exactly 500 mL of the water. Calculate the molar concentration of Pb2* if 0.00275 g of NazSO4 was needed for the complete precipitation of lead ions as Pbso4 Enter your answer in the provided box. The concentration of lead ions (Pb2) in a sample of...
Solubility product constant Ksp is also important to predict whether the precipitation will occur under the known condition of ion concentration. This can be done by comparing the solubility quotient (Q) with the value of Ksp of the tested compound. 12. A solution of 0.00016 M lead (II) nitrate, or Pb(NO3)2, was poured into 450 mL of 000023 M sodium sulfate, Na2SO4. Would a precipitate of lead(II)sulfate, PbSO4, be expected to form if 250 mL of the lead nitrate solution...
12. A solution of 0.00016 M lead (II) nitrate, or Pb(NO3)2, was poured into 450 mL of 0.00023 M sodium sulfate, Na2S04. Would a precipitate of lead(II)sulfate, PbSO4, be expected to form if 250 mL of the lead nitrate solution were added? Write the equilibrium equation of PbSO4 dissociation in water.
Review | Constants Periodic Table MISSED THIS? Read Section 5.6 (Pages 183 - 185). Part B Lead ions (Pb2+) can be removed from solution by precipitation with SO42-. Suppose that a solution contains aqueous Pb(NO3)2. Enter the net ionic equation to show the reaction of aqueous Pb(NO3)2 with aqueous sodium sulfate to form solid PbSO4 and aqueous sodium nitrate. Express your answer as a chemical equation. Identify all of the phases in your answer. ► View Available Hint(s) - AEV...
3. A1.22 g solid sample of impure lead (11) nitrate was dissolved in 40.0 mL of distilled water and analyzed by gravimetric analysis. Sodium sulfate was delivered from a burette to precipitate the lead ions as lead sulfate. Sodium sulfate was added until no more precipitate was seen to form. The precipitate was filtered, washed an dried before been weighed. When weighed, the precipitate had a mass of 1.44 g. What is the mass percent of lead in the original...
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A 1.04 g sample of KBr is dissolved in water to give 155 mL of solution. This solution is then added to 165 mL of 0.015 M aqueous Pb(NO3)2, in an attempt to remove the toxic lead(II) ions from the solution via precipitation as insoluble PbBr2(s). The precipitation reaction that occurs is: Pb2+ (aq) + 2 Br (aq) ---> PbBr2 (s) At the end of the reaction, what is the concentration (in molarity) of nitrate ions in the solution? Note:...
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To test a water sample for lead, 25 grams of sodium chloride is dissolved into 100.0 mL of the water. If no precipitation occurs, the concentration of lead(II) ions in the water is below what concentration?
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