Answer:
Mass of precipitate (PbSO₄) formed = 0.696 g
Explanation:
The reaction of Lead II Nitrate and Sodium Sulfate is given as
Pb(NO₃)₂ (aq) + Na₂SO₄ (aq) → 2NaNO₃ (aq) + PbSO₄ (s)
The precipitate from this reaction is the PbSO₄
0.76 g of Pb(NO₃)₂ is available for reaction, we convert to number of moles
Number of moles = (Mass)/(Molar Mass)
Mass = 0.76 g
Molar mass of Pb(NO₃)₂ = 331.2 g/mol
Number of moles of Pb(NO₃)₂ = (0.76/331.2) = 0.002294686 moles = 0.002295 moles
25.00 ml of 0.2010M sodium sulfate is available for the reaction, we also conert to number of moles.
Number of moles = (Concentration in mol/L) × (Volume in L)
Concentration of Na₂SO₄ in mol/L = 0.201 M
Volume of Na₂SO₄ in L = (25/1000) = 0.025 L
Number of moles of Na₂SO₄ = 0.201 × 0.025 = 0.005025 moles
From the stoichiometric balance of the reaction, 1 mole of Pb(NO₃)₂ reacts with 1 mole of Na₂SO₄
Hence, this indicates that Pb(NO₃)₂ is the limiting reagent as it is in short supply and it will therefore dictate the amount of precipitate (PbSO₄) formed.
1 mole of Pb(NO₃)₂ gives 1 mole of the precipitate (PbSO₄)
0.002295 mole of Pb(NO₃)₂ will give 0.002295 mole of the precipitate (PbSO₄)
Mass = (Number of moles) × (Molar mass)
Number of moles of precipitate (PbSO₄) formed = 0.002295 mole
Molar mass of PbSO₄ = 303.26 g/mol
Mass of precipitate (PbSO₄) formed = 0.002295 × 303.26 = 0.6958864734 = 0.696 g
Hope this Helps!!!
solubility will increase, since it is basic in nature, on acidifying solubility increases.
being basic on acidification, the solubility increases.
is alkaline in nature, it gets neutralized and there is no increase in acidity.
is a salt and it shows no effect on acidification.
is a salt but insoluble in water and shows no effect on acidification.
is strongly basic in nature, on acidifying the solubility increases.