A) \[\text{64}\,\text{mL}\,\text{1M}\,\text{NaOH}\,\text{+}\,\text{33}\,\text{mL}\]of \[\text{0}\text{.5}\,\text{M}\,{{\text{H}}_{\text{2}}}\text{S}{{\text{O}}_{\text{4}}}\]
B) \[\text{33}\,\text{mL}\]of \[\text{1}\,\text{M}\,\text{NaOH}\,\text{+}\,\text{67}\,\text{mL}\,\]of \[\text{0}\text{.5}\,\text{M}\,{{\text{H}}_{\text{2}}}\text{S}{{\text{O}}_{\text{4}}}\]
C) \[\text{40}\,\text{mL}\]of \[\text{1}\,\text{M}\,\text{NaOH}\,\text{+}\,\text{60}\,\text{mL}\]of \[\text{0}\text{.5}\,\text{M}\,{{\text{H}}_{\text{2}}}\text{S}{{\text{O}}_{\text{4}}}\]
D) \[\text{50}\,\text{mL}\]of \[\text{1}\,\text{M}\,\text{NaOH}\,\text{+}\,\text{50}\,\text{mL}\]of \[\text{0}\text{.5}\,\text{M}\,{{\text{H}}_{\text{2}}}\text{S}{{\text{O}}_{\text{4}}}\]
Correct Answer: B
Solution :
The combination, in which concentration of \[{{\text{H}}^{\text{+}}}\] is maximum, will produce the highest rise in temperature. Conc. of \[O{{H}^{-}}=\frac{67\times 1-33\times 1}{67+33}\] [\[\because \,N=2M\]for \[{{\text{H}}_{\text{2}}}\text{S}{{\text{O}}_{\text{4}}}\]] \[=0.34=3.4\times {{10}^{-1}}\] \[[{{H}^{+}}][O{{H}^{-}}]=1\times {{10}^{-14}}\] \[\therefore \] \[[{{H}^{+}}]=\frac{1\times {{10}^{-14}}}{3.4\times {{10}^{-1}}}=2.94\times {{10}^{-14}}\] \[[{{H}^{+}}]=\frac{67\times 1-33\times 1}{67+33}=0.34\] \[[{{H}^{+}}]=\frac{60\times 1-40\times 1}{60+40}=0.20\] Since concentration and volume of acid and base is equal, the solution is neutral ie, \[[{{H}^{+}}]={{10}^{-7}}\] Hence, mixture given in option will produce the highest rise in temperature.
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