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Grade 11Inorganic Chemistry

A mixture of 100 m mol of Ca(OH)2 and 2g of sodium sulphate was dissolved in water and the volume was made up to 100 mL. The mass of calcium sulphate formed and the concentration of OH- in resulting solution are
given that molar mass of CaSO4=136 , Ca(OH)2=74 and that of Na2SO4 = 143 and also KSP of Ca(OH)2 is 5.5x10-6

Profile image of preet kaur
7 Years agoGrade 11
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3 Answers

Profile image of Mayank
7 Years ago
 Ca(OH)2  + Na2SO4 ----------> CaSO4 + 2NaOH
100 m mol   14 m mol                   -------      ------
    --------        --------                   14 m mol     28 m mol

WCaSO4=
[OH–] = 
28
0.28M
100
 
Profile image of Arun
ApprovedApproved Tutor Answer7 Years ago

Na2SO4 + Ca(OH)2 → CaSO4 + 2 NaOH

 mmol of Na2SO4 = 2*1000/143

                          =13.98 m Mol

mmol of CaSO4 formed = 13.98 m Mol 

Mass of CaSO4 formed = 13.98 × 10^-3 × 136 = 1.90 g

mmol of NaOH = 28 mmol

Ca(OH)2 → Ca2+ + 2 OH-

[OH–] = 28 mmol / 100ml=0.028 mol/0.1 L =0.28mol L-1

Profile image of Mayank
ApprovedApproved Tutor Answer7 Years ago
 
 
 
 
 
 
Ca(OH)2  + Na2SO4 ----------> CaSO4 + 2NaOH    (i am sorry but the above ans is incomplete)
100 m mol   14 m mol                   -------      ------
    --------        --------                   14 m mol     28 m mol
WCaSO4=14x10-3x136=1.9gm
[OH-]=28/100=0.28M
Hence the ans................

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