A) 7
B) > 7
C) < 7
D) none of these
Correct Answer: B
Solution :
Given, NaOH V = 10 mL, N = 0.1 \[{{H}_{2}}S{{O}_{4}}V=10\,\,mL\,\,N=0.05\] Milli equivalents of \[NaOH=10\times 0.1=1\] Milli equivalents of \[{{H}_{2}}S{{O}_{4}}=10\times 0.05=0.05\] \[\because \] 0.05 milli equivalents of \[{{H}_{2}}S{{O}_{4}}\] neutralises 0.05 milli equivalents of NaOH. \[\therefore \] 0.05 milli equivalents of NaOH will remain in solution. Solution is basic. Basic solution has \[pH>7\]. pH of solution is more than 7.
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