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