A) less than\[7\]
B) \[7\]
C) zero
D) greater than\[7\]
Correct Answer: B
Solution :
Given, for\[NaOH,\,\,V=10\,\,mL,\,\,N=0.1\,\,N\] For\[{{H}_{2}}S{{O}_{4}},\,\,V=10\,\,mL,\,\,N=0.05\,\,N\] Miliequivalents of\[NaOH=10\times 0.1=1\] Miliequivalents of\[{{H}_{2}}S{{O}_{4}}=10\times 0.05=0.5\] \[\underset{1\,\,equivalent}{\mathop{{{H}_{2}}S{{O}_{4}}}}\,+\underset{2\,\,equivalent}{\mathop{2NaOH}}\,\xrightarrow{{}}N{{a}_{2}}S{{O}_{4}}+{{H}_{2}}O\] \[\because \]\[0.5\] equivalent of \[{{H}_{2}}S{{O}_{4}}\] will react with 1 equivalent of\[NaOH\] \[\therefore \]The pH of solution\[=7\] (neutral)
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