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`CuSO_4` crystallise in rock salt structure.Its cell parameter can be determined by various experimental methods like electrical conductivity measurement, colligative properties measurement, pH measurement etc. A cubic crystal of `CuSO_4` of edge length 17.1 mm, is dissolved in water to make 500 ml solution of pH 5.`root3(1.5)=1.14)` Given : `Cu(H_2O)_6^(2+)+H_2O hArr [Cu(H_2O)_5(OH)]^(+)+H_3O^(+) , K=10^(-5)` In the given solution is made 1 M with respect to `[Cu^(2+)]` becomes `10^(-15)` M therefore `K_f` for the formation of `Cu(NH_3)_4^(2+)` isA. `10^10`B. `10^8`C. `2xx10^8`D. `2xx10^10` |
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Answer» Correct Answer - D `{:(Cu(H_2O)_6^(2+)+H_2OhArr,[Ca(H_2O)_5(OH)]^(+),+H_3O^(+)),(C, , ),(C-x,x,x):}` `10^(-5)=x^2/(C-x)=((10^(-5))^2)/(C-10^(-5))` `C=2xx10^(-5)` mol/lt `:.` moles of `CuSO_4` dissolved =`2xx10^(-5)xx0.5 =10^(-5)` Number of units cells `=(10^(-5)xxN_A)/4=6/4xx10^(18)=1.5xx10^(18)` number of unit cells along one edge of the cube =`root3(1.5xx10^(18))=1.14xx10^8` If edge length of F C C unit cell is a Now`{:(Cu^(2+)+,4NH_3hArr,Cu(NH_3)_4^(2+)),(2xx10^(-5), , ),(10^(-15),1,2xx10^(-5)):}` `K_f=([Cu(NH_3)_4^(2+)])/([Cu^(2+)][NH_3]^(4))=(2xx10^(-5))/(10^(-15)xx1)=2xx10^10` |
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