a)Â Â Â Assuming 100% dissociation of LiOH
[H3O+]  = 10-14 / [8.86×10−3 ]
= 1.128 x 10 -12
[OH−]    =
8.86×10−3                                     Â
pH,  = -log (1.128 x 10 -12) = 12 - 0.052
= 11.947
pOH = 14 - pH = 14 - 11.947 = 2.053
B)
a)Â Â Â Assuming 100% dissociation of
Ba(OH)2
[H3O+]Â Â = 10-14 / [2x
1.14×10−2] = 4.38 x 10-13
[OH−]    = 2.28 X 10-2
                                   Â
pH,  = -log (4.38 x 10-13) = 13 - 0.64 =
12.3
pOH = 14 - pH = 14 - 12.3 = 1.64
Please post the remaining separately or solve as above...
they are all similar.....
