A) 67 mL of 1 M NaOH + 33 mL of 0.5 M \[{{H}_{2}}S{{O}_{4}}\]
B) 33 mL of 1 M NaOH + 67 mL of 0.5 M \[{{H}_{2}}S{{O}_{4}}\]
C) 40 mL of 1 M NaOH + 60 mL of 0.5 M \[{{H}_{2}}S{{O}_{4}}\]
D) 50 mL of 1 M NaOH + 50 mL of 0.5 M \[{{H}_{2}}S{{O}_{4}}\]
Correct Answer: B
Solution :
The combination, in which concentration of \[{{H}^{+}}\] is maximum, will produce the highest rise in temperature. [a] Cone. of \[O{{H}^{-}}=\frac{67\times 1-33\times 1}{67+33}\] [\[\because \,N=2\,M\] for \[{{H}_{2}}S{{O}_{4}}\]] \[=0.34=3.4\times {{10}^{-1}}\] \[[{{H}^{+}}]\,[O{{H}^{-}}]=1\times {{10}^{-14}}\] \[\therefore \] \[[{{H}^{+}}]=\frac{1\times {{10}^{-14}}}{3.4\times {{10}^{-1}}}=2.94\times {{10}^{-14}}\] [b] \[[{{H}^{+}}]=\frac{67\times 1-33\times 1}{67+33}=0.34\] [c] \[[{{H}^{+}}]=\frac{60\times 1-40\times 1}{60+40}=0.20\] [d] Since concentration and volume of acid and base is equal, the solution is neutral ie,\[[{{H}^{+}}]={{10}^{-7}}\] Hence, mixture given in option [b] will produce the highest rise in temperature.
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