A) \[1.11\times {{10}^{-4}}M\]
B) \[3.7\times {{10}^{-4}}M\]
C) \[3.7\times {{10}^{-3}}M\]
D) \[1.11\times {{10}^{-3}}M\]
Correct Answer: B
Solution :
Key Idea: \[[{{H}^{+}}]\]in mixture \[[{{H}^{+}}]\] of 1st acid x its volume \[+[{{H}^{+}}]\]of IInd acid \[\times \] its volume \[+[{{H}^{+}}]\] of\[=\frac{IIIrd\,acid\times its\,volume}{Total\,volume}\] \[M=\frac{{{M}_{1}}{{V}_{1}}+{{M}_{2}}{{V}_{2}}+{{M}_{3}}{{V}_{3}}}{V}\] Assume the volume of each solution is 1 L. \[[{{H}_{3}}{{O}^{+}}]\]in solution of pH = 3 is \[{{10}^{-3}}M\] \[[{{H}_{3}}{{O}^{+}}]\]in solution of pH = 4 is \[{{10}^{-4}}M\] \[[{{H}_{3}}{{O}^{+}}]\] in solution of pH = 5 is \[{{10}^{-5}}M\] Total \[[{{H}_{3}}{{O}^{+}}]=\left( \frac{{{10}^{-3}}+{{10}^{-4}}+{{10}^{-5}}}{3} \right)\]\[=0.00037\] \[=3.7\times {{10}^{-4}}M\] Note: If mixture contains both acids and bases, then concentration of mixture is given as Concentration of mixture cone of acids\[\times \] their volumes - \[=\frac{\text{cone of bases}\times \text{their volumes}}{\text{Total}\,\text{volumes}}\]
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