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author | priyanka | 2015-06-24 15:03:17 +0530 |
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committer | priyanka | 2015-06-24 15:03:17 +0530 |
commit | b1f5c3f8d6671b4331cef1dcebdf63b7a43a3a2b (patch) | |
tree | ab291cffc65280e58ac82470ba63fbcca7805165 /821/CH7/EX7.27/7_27.sce | |
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Diffstat (limited to '821/CH7/EX7.27/7_27.sce')
-rwxr-xr-x | 821/CH7/EX7.27/7_27.sce | 9 |
1 files changed, 9 insertions, 0 deletions
diff --git a/821/CH7/EX7.27/7_27.sce b/821/CH7/EX7.27/7_27.sce new file mode 100755 index 000000000..b134a51f9 --- /dev/null +++ b/821/CH7/EX7.27/7_27.sce @@ -0,0 +1,9 @@ +Ksp=2.2*10^-8;//Solubility product of PbSO4//
+Pb=0.01;//concentration of Pb in Pb(NO3)2//
+SO4=Ksp/Pb;//Concentration of SO4 in PbSO4 solution//
+printf('Let us first calculate the maximum concentration of SO4 that can remain in equilibrium with PbSO4 if the concentration of Pb is 0.01M');
+printf('\nConcentration of SO4 in PbSO4 solution=Ksp=2.2*10^-6M\nThe concentration of SO4 should be greater than 2.2*10^-4M in order to precipitate Pb from a 0.01M solution as PbSO4');
+Pb2=Ksp/0.001;
+printf('\nConcentration of Pb in PbSO4 solution=Pb2=2.2*10^-5mol per litre');
+printf('\nHence out of 0.01moles of Pb in a litre only 2.2*10^-5mol per litre remain in solution.the precipitation is almost 99.78percent complete.');
+printf('\nTherefore it can be said that Pb is quantitatively precipitated in these conditions.');
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