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