A) 2.3
B) 1.9
C) 4.3
D) 5.1
Correct Answer: B
Solution :
\[pH\,of\,A=3,\,pH\,of\,B=2\] \[As\,[{{H}^{+}}]={{10}^{-3}}M,\,[{{H}^{+}}]={{10}^{-2}}\,M\] Total \[[{{H}^{+}}]={{10}^{-3}}+{{10}^{-2}}={{10}^{-3}}+10\times {{10}^{-3}}\] \[=11\times {{10}^{-3}}\] So \[pH=-{{\log }_{10}}[{{H}^{+}}]=-\log (11\times {{10}^{-3}})\] \[=3-\log 11\] \[=3-1.04=1.96\] Hence, the correct option is [b].You need to login to perform this action.
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