A) \[3-log\text{ }2.8\]
B) \[7\text{ }-log\text{ }2.8g\]
C) \[4-log\text{ }2.8\]
D) \[5-log\text{ }2.8\]
Correct Answer: C
Solution :
: \[pH=3\]means \[{{10}^{-3}}M\]and \[pH=4\]means \[{{10}^{-4}}M\]. \[100mL\]of \[{{10}^{-3}}M\]solution contain \[=\frac{{{10}^{-3}}\times 100}{1000}=10\times {{10}^{-5}}mol\] \[400mL\]of \[{{10}^{-4}}M\]solution contain \[=\frac{{{10}^{-4}}\times 400}{1000}=4\times {{10}^{-5}}mol\] Total number of moles present \[=(10\times {{10}^{-5}})+(4\times {{10}^{-5}})=14\times {{10}^{-5}}mol\] Total volume after mixing \[=(100+400)=500mL\] Molarity of the mixture \[=\frac{14\times {{10}^{-5}}}{500}\times 1000\] \[=\left( \frac{14}{5}\times {{10}^{-4}} \right)M\] pH of the mixture solution \[=-\log \left( \frac{14}{5}\times {{10}^{-4}} \right)\] \[=-\log 2.8-(-4)\] \[pH=4-\log 2.8\]
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