In the lab, various concentrations of Ca2+were determined. The solutions were pr

Question

In the lab, various concentrations of Ca2+were determined. The solutions were prepared using solid calcium iodate that was added in excess to a 0.015 M calcium nitrate solution and allowed to equilibrate. To determine the iodate ion concentration of one solution, 25.00 ml of the calcium iodate solution was titrated with 30.10 ml of a 0.150 M potassium thiosulfate solution. Using this information, and balanced chemical equation below, what is the final Ca2+ concentration in the solution at equilibrium?

Explanation / Answer

Determination of iodate ion concentration:

The reaction between thiosulfate and iodate(Iodometric titration) is given by

IO3-(aq) +5I- (aq) +6H+(aq)3I2 (aq) +3H2O(l)

2S2O32- (aq) +I2(aq) S4O62-(aq) +2I-(aq)]*3

--------------------------------------------------------------------------

net eqn:IO3-(aq) +6S2O32- (aq)+6H+(aq)3S4O62-(aq)+I- (aq) +3H2O(l)

So,IO3-(aq) reacts with S2O32- (aq) in 1:6 molar ratio.

[IO3-]/[S2O32-]=1/6

6[IO3-]=[S2O32-]

Using Equation,

V(thio)*M(thio)=6*V(iodate)*M(iodate)

So, M(iodate)=V(thio)*M(thio)/6*V(iodate)=30.10ml*0.150M/6*25ml= 0.0301M

calcium iodate that was added in excess so that it saturates the solution,

Ca(IO3)2 (s)<--->Ca2+(aq)+ 2IO3-(aq) ,Ksp=6.5*10^-6

Let the Solubility of Ca2+ be S ,then [IO3-]=2S

Ksp=6.5*10^-6=[Ca+][IO3-]^2=S(2S)^2

So, 6.5*10^-6=4S^3

S=0.0117 M

Equilibrium existing in solution of Ca(NO3)2 +Ca(IO3)2

Ca(IO3)2 (s)<--->Ca2+(aq)+ 2IO3-(aq) ,Ksp=6.5*10^-6

Ca(NO3)2 (s) --->Ca2+(aq)+ 2NO3-(aq) (soluble,so complete dissociation)

[Ca2+]eq=[Ca(NO3)2]=0.0150M

Calcium nitrate has more solubility than calcium iodate so the eqm will shift in reverse direction ,due to excess of Ca2+ ions, and thus [Ca2+]eq=[Ca(NO3)2]=0.0150M

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