0.1 millimole of CdSO4 is present in 10mL acid solution of 0.08 N HCl.Now H2S is passed to ppt. all the Cd2+ ions.The pH of the soln after filtering off ppt,boiling off H2S and making the solution 100 mL by adding water is approx --
A. 2
B.4
C.6
D.8
11It CdSo4 + H2S = CdS + H2so4
There are 0.1 milimoles of So4 Thereofere increase in H+ will be 0.2 millimoles
HCl is 0.08N therefore molarity of H is 0.08M
10 ml Therefore 0.0008Moles
Total H+ = 0.001moles
NOw in 100ml The concentration is 0.01 M Hence pH = 2
Virang your solution was of great help!!! But I would like to bring to your notice that you have made a grave mistake in the last step.
Before dilution volume is 10ml and concentration of H+ is .001 moles. So after solution to 100ml the H+ concentration will be. 0001 ...therefore the answer should be rather 4.
