A) \[8\]
B) \[2\]
C) \[1\]
D) \[3\]
Correct Answer: B
Solution :
Total mill equivalents of \[{{H}^{+}}\] \[=30\times \frac{1}{3}+20\times \frac{1}{2}=20\] Total mill equivalents of \[O{{H}^{-}}\] \[=40\times \frac{1}{4}=10\] Milli equivalence of \[{{H}^{+}}\]left \[=20-10=10\] \[\therefore \] \[[{{H}^{+}}]=\frac{10}{1000}g\,ions/d{{m}^{3}}={{10}^{-2}}\] \[\therefore \] \[pH=2\]You need to login to perform this action.
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