A) 2
B) 3
C) 2.5
D) 2.8
Correct Answer: C
Solution :
100 mL of a \[HCl\] solution of \[pH=2\Rightarrow [{{H}^{+}}]={{10}^{-2}}\] 400 mL of a \[HCl\] solution of \[pH=3\Rightarrow [{{H}^{+}}]={{10}^{-3}}\] Resulting \[[{{H}^{+}}]=\frac{100\times {{10}^{-2}}+400\times {{10}^{-3}}}{500}\]\[=\frac{1+0.4}{500}=\frac{1.4}{500}=0.0028\] \[\because \] \[pH=-\log [{{H}^{+}}]\] \[\therefore \] \[pH=-\log (0.0028)=2.55\]You need to login to perform this action.
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