diff options
Diffstat (limited to '821/CH7/EX7.31/7_31.sce')
| -rwxr-xr-x | 821/CH7/EX7.31/7_31.sce | 11 |
1 files changed, 11 insertions, 0 deletions
diff --git a/821/CH7/EX7.31/7_31.sce b/821/CH7/EX7.31/7_31.sce new file mode 100755 index 000000000..787b06b78 --- /dev/null +++ b/821/CH7/EX7.31/7_31.sce @@ -0,0 +1,11 @@ +C=0.01;//concentration of Ca(NO3)2 solution//
+Ksp=3.2*10^-11;//Solubility product of Fe(OH)3//
+printf('CaF2 = Ca2+ + 2F-\n(Ca2+)(F-)^2 = 4*S^3 = 3.2*10^-11.');
+printf('\nLet S1 be the solubility in 0.01M Ca(NO3)2\nCa(NO3)2 can be assumed to dissociate completely so that (Ca2+) from Ca(NO3)2 is 0.01M');
+S=(Ksp/4)^0.33;//solubility in mol per litre//
+printf('\nSolubility of CaF2 solution=S=%f=2.18*10^-4mol per litre',S);
+printf('\nThe solubility product relationship should be true,irrespective of the source Ca2+\nCompared to the concentration of Ca2+ ions obtained from Ca(NO3)2,that of Ca2+ ions from CaF2 is negligible');
+S1=sqrt(Ksp/0.04);//solubility in 0.01M ca(NO3)2//
+printf('\nBut the F- ions are obtained only from CaF2 and so (F-)=2*S1\nKsp = 3.2*10^-11=(S1+0.01)*(2*S1)^2=(0.01)*(2*S1)^2 since S1 is negligible compared to 0.01');
+printf('\nSolubility in 0.01M Ca(NO3)2 solution=S1=%f=2.83*10^-5',S1);
+printf('\nThus the value of S1 can be seen to be less than that of S');
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