A) If 2.0 litre of a solution of\[{{H}_{2}}S{{O}_{4}}\]contains 0.1 mole, then pH of the solution is 2
B) The concentration of\[O{{H}^{-}}\]in 0.005 M\[HN{{O}_{3}}\]is \[2.0\times {{10}^{-12}}mol/1\]
C) The pH of 0.01 M KOH is 12
D) In a 0.001 M solution of\[NaOH\]the concentration of\[{{H}^{+}}\]is\[{{10}^{-3}}\text{ }mol/litre\]
Correct Answer: D
Solution :
In\[0.001\text{ }M\text{ }NaOH,\] \[[O{{H}^{-}}]=0.001={{10}^{-3}}\] \[\therefore \] \[[{{H}^{+}}]=\frac{{{10}^{-14}}}{[O{{H}^{-}}]}=\frac{{{10}^{-14}}}{{{10}^{-3}}}={{10}^{-11}}\] Hence, [d]. is incorrect.
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