A) 4.74
B) 2.26
C) 9.26
D) 5.00
Correct Answer: C
Solution :
[c] 9.26 \[40\,mL\,of\,0.1\,MN{{H}_{3}}\,solution\,=40\times 0.1\] \[=4\,mil\operatorname{limole}\] \[20\,ml\,of\,0.1\,MHCl\,solution\,=20\times 0.1\] \[=2\,mil\operatorname{limole}\] \[N{{H}_{4}}OH+HCl\to N{{H}_{4}}Cl+{{H}_{2}}O\] 2 millmole of HCI will neutralize 2 millmoles of \[N{{H}_{4}}OH\] to form 2 millimoles of \[N{{H}_{4}}Cl\] \[N{{H}_{4}}OH\,left\,=\,2\,mil\lim oles\] \[Total\,vlume\,=60ml\] \[\therefore \,[N{{H}_{4}}OH]=\frac{2}{60}H[N{{H}_{4}}Cl]=\frac{2}{40}M\] \[pOH=p{{K}_{b}}+\frac{\log \,[N{{H}_{4}}Cl]}{[N{{H}_{4}}OH]}\] \[=4.74+\log \frac{2/60}{2/60}=4.74\] \[PH=14-4.74=9.26\]
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