A) 3.2
B) 1.9
C) 3.4
D) 3.5
Correct Answer: B
Solution :
pH of A = 3 pH of B = 2 \[\therefore \] \[[{{H}^{+}}]={{10}^{-3}}\,M,\] \[[{{H}^{+}}]={{10}^{-2}}\,M\] Total \[[{{H}^{+}}]\] \[={{10}^{-3}}+{{10}^{-2}}={{10}^{-3}}+10\times {{10}^{-3}}\] \[=11\times {{10}^{-3}}\] \[\therefore \] \[pH=-{{\log }_{10}}[{{H}^{+}}]=-\log \,(11\times {{10}^{-3}})\] \[=3-\log \,11\] \[=3-1.04=1.96\]You need to login to perform this action.
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