A) 8
B) 2
C) 1
D) 3
Correct Answer: B
Solution :
Total milli equivalents of \[{{H}^{+}}\] \[=30\times \frac{1}{3}+20\times \frac{1}{2}=20\] Total milli 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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