A) 5
B) 10.4
C) 12.3
D) 1.7
Correct Answer: C
Solution :
Number of mill equivalents of \[\operatorname{NaOH}= 800\,\,\times \,\,0.05 =40\] Number of mill equivalents of HCI \[=\,\,200\times 0.1 =20\] \[\operatorname{Total} volume = 200 + 800 = 1000 mL\] Number of mill equivalents of NaOH present in \[40-20=20\] \[\operatorname{pOH}=-log{{\left[ OH \right]}^{-}}\] \[=-\,log\left[ 2\,\,\times \,\,1{{0}^{-}}^{2} \right]\] \[=\,\,\,1.7\] also \[\operatorname{pH}\,\,+\,\,pOH=14\] \[\operatorname{pH}=14-pOH\] \[=\,\,\,14-1.7=12.3\]You need to login to perform this action.
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