# Determine the solubilities of silver chromate, barium chromate, ferric, hydroxide, lead, chloride and mercurous iodide at 298 K. Given K_(sp) values

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Determine the solubilities of silver chromate, barium chromate, ferric, hydroxide, lead, chloride and mercurous iodide at 298 K. Given K_(sp) values : Ag_(2)CrO_(4)=1.1xx10^(-12), BaCrO_(4)=1.2xx10^(-10), Fe(OH)_(3)=1.0xx10^(-38), PbCl_(2)=1.6xx10^(-5), Hg_(2)I_(2)=4.5xx10^(-29)
Determine also the molarities of individual ions.

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Correct Answer - S[BaCrO_(4)]=1.1xx10^(-5)M=[Ba^(2+)]=[CrO_(4)^(2-)]
S(Fe(OH)_(3))=1.39xx10^(-10)M=[Fe^(3+)], [OH^(-)]=3xx1.39xx10^(-10) M = 4.17xx10^(-10)M
S(PbCl_(2))=1.59xx10^(-2)M=[Pb^(2+)],[Cl^(-)]=2xx1.59xx10^(-2)M=3.18xx10^(-2)M
S(Hg_(2)I_(2))=2.24xx10^(-10)M=[Hg_(2)^(2+)], [I^(-)]=2xx2.24xx10^(-10)M=4.48xx10^(-10)M
{:(Ag_(2)CrO_(4) hArr 2"Ag"^(+)+CrO_(4)^(2-),K_(sp)=(2s)^(2)(s)=4s^(3)"," :. s^(3)=K_(sp)//4,,),(s" " 2s" " s" "=(1.1xx10^(-12))//4=0.275xx106(-12)=2.75xx10^(-13),,):}
:. 3 log s = log (2.75xx 10^(-13))=-13+0.4393=-12.5607
or  log s = - 4.1869 = bar(5).8131
:. s= 6.5xx10^(-5)  mol L^(-1)
[Ag^(+)]=2xx6.5xx10(-5)=13.0xx10^(-5)M = 1.30 xx 10^(-4)M
[CrO_(4)^(2-)] = 6.5xx10^(-5)M
For other salts, proceed exactly in similar manner .