Hydrochemistry Practical Handbook

HYDRO CHEMISTRY

Hydro Chemistry Practical Hand Book

PRACTICALS 4 AND 5

Question 1 . Calculation of Solubility Product of a compound , from the given Solubility data

Principle . The product of the Ionic concentrations , in gram mole per litre , of a saturated solution of the sparingly soluble salt is nearly equal to the Solubility Product . Pl . follow the steps

Calculate the molecular weight of the compound from the given data of atomic weight

Convert the given solubility data in terms of gm / L

Calculate solubility data in terms of molar concentration

- gm mole / L This is got by dividing solubility data - gm /l , in step 2 by molecular weight - gm /gm mol , i .e . the data got in step 1 . This gives the concentrations of respective ions also

See how many ions (element or radical ) can be formed from the compound say two or three ions depending upon the compound and calculate the Solubility product by taking the products of all the ion concentrations

Example : Sphalerite , Zn S

The molecular weight of Zn S 65 .39 32 .066 97 .456 gm /gm mol

Solubility 0 .00065g /1000ml [ (0 .00065g /1000ml / 1000ml / L] 0 .00065 g / L

Molar concentration of sat . solution of ZnS (0 .00065 g / L (97 .456 gm /gm mol 0 .0000066 gm mol / L . So , conc . of Zn Conc . of S 0 .0000066 gm mol / L

S .P of Zn S [Zn]x [S] 0 .0000066 x 0 .0000066 4 .356x10 - 11

Similarly the Solubility products of others...