before
dilution                                     Â
after dilution
M1 =
1MÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â
M2
V1 =
25ml                                               Â
V2 = 750ml
          Â
M1V1Â Â Â = M2V2
           Â
M2Â Â Â = M1V1/V2
                    Â
= 1*25/750 = 0.034M
HCl + H2O---------> H3O+ + Cl-
0.034 MÂ Â Â Â Â Â 0.034M
[H3O+] = [HCl]
[H3O+]Â Â = 0.034M
PH = -log[H+]
        =
-log0.034Â Â = 1.4685
Ba(OH)2 + 2HCl --------> BaCl2 + H2O
no of moles of Ba(OH)2 = W/G.M.Wt = 0.35/171.34 =
0.00204moles
Ba(OH)2 + 2HCl --------> BaCl2 + H2O
1 moles of ba(OH)2 react with 2 moles of HCl
0.00204 moles of Ba(OH)2 react with = 2*0.00204/1Â Â =
0.00408moles of HCL
molarity = no of moles/volume in L
0.034Â Â Â Â Â = 0.00408 /volume in L
volume in L = 0.00408/0.034 = 0.12L = 120ml >>>
answer
