A) \[1.6\]
B) \[12.4\]
C) \[13.4\]
D) \[10.8\]
Correct Answer: B
Solution :
Concentration of mixture mill equivalents of \[NaOH\] \[=\frac{-milliequivalents\,of\,HCl}{total\,volume}\] \[=\frac{V\times \frac{1}{10}-V\frac{1}{20}}{(V+V)}\] \[=\frac{V(0.1-0.05)}{2V}\] \[=0.025\] Since, \[NaOH\] is in excess, \[[O{{H}^{-}}]=0.025\] \[pOH=1.6\] \[pH=14-1.6=12.4\]You need to login to perform this action.
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