A) The conjugate base of \[{{H}_{2}}P{{O}_{4}}^{-}\] is \[HP{{O}_{4}}^{2-}.\]
B) pH + pOH = 14 for all aqueous solutions.
C) the pH of \[1\times {{10}^{-8}}M\,HCL\] is 8.
D) 96,500 coulombs of electricity when passed through a \[CuS{{O}_{4}}\] solution deposits 1 gram equivalent of copper at the Cathode.
Correct Answer: C
Solution :
[c] The pH of \[1\times {{10}^{-8}}M\]HCl is always less than 7. Since the solution is of acid. It can be shown as \[[{{H}^{+}}]={{10}^{-7}}\] from \[{{H}_{2}}O\] and \[{{10}^{-8}}\] from HCl \[[{{H}^{+}}]={{10}^{-8}}(10+1)\] \[[{{H}^{+}}]={{10}^{-8}}\times 11\] \[-\log [{{H}^{+}}]=8\,\log 10-\log 11\] \[pH=8-\log 11\] \[pH=6.93.\]
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