Lead sulfate. PbSO_4, was used as a white paint pigment. When a solution is prep
ID: 518433 • Letter: L
Question
Lead sulfate. PbSO_4, was used as a white paint pigment. When a solution is prepared that is 5.0 times 10^-3 M in lead ion. Pb^2+ and 1.0 times 10^-4 M in sulfate ion. SO^2-_4, would you expect some of the lead sulfate to precipitate? The following solutions are mixed: 1.0 L of 0.00010 M NaOH and 1.0 L of 0.0014 M MgSO_4. Isa precipitate expected? Explain. A45-mLsampieof 0.015 M calcium chloride, CaCl_2, is added to 55 mL of 0.010 M sodium sulfate. Na_2SO_4. Is a precipitate expected? Explain. A 45.0-mL sample of 0.0015 M BaCl_2 was added to a beaker containing 75.0 mL of 0.0025 M KF. Will a precipitate form? A 65.0-mL sample of 0.010 M Pb(NO_3)_2 was added to a beaker containing 40.0 mL of 0.035 M KCl. Will a precipitate form? How many moles of calcium chloride, CaCl_2, can be added to 1.5 L of 0.020 M potassium sulfate, K_2SO_4, before a precipitate is expected? Assume that the volume of the solution is not changed significantly by the addition of calcium chloride. Magnesium sulfate, MgSO_4, is added to 456 mL of 0.040 M sodium hydroxide, NaOH, until a precipitate just forms. How many grams of magnesium sulfate were added? Assume that the volume of the solution is not changed significantly by the addition of magnesium sulfate.Explanation / Answer
1.0 L of 0.0001M Na OH is mixed with 1.0L of 0.00 14 M Mg SO4. precipitate is expected, explain why?write out the solution.
the equation is
2 NaOH + MgSO4 >> Na2SO4 + Mg(OH)2
Ksp = 1.8 x 10^-11
Moles OH- = 1.0 x 0.0001 = 0.00001
Moles Mg2+ = 1.0 x 0.0014 = 0.0014
Total volume = 2.0 L
Concentration OH- = 0.00001 M
Concentration Mg2+ = 0.0014 M
=0.0014 x ( 0.00001)^2 = 1.96 x 10^-16 .
Since this ion-product is not appreciably less than Ksp hence may expect precipitation
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.