A) 1 : 1
B) 1 : 2
C) 2 : 1
D) 3 : 1
Correct Answer: C
Solution :
pH = 9.26 shows that \[[N{{H}_{4}}OH]\,\,>\,HCl\] \[\therefore \] The mixture is a buffer. Let \[HCl=x\,\,mL=x\,\,mmol\] \[\therefore \] \[N{{H}_{4}}OH=\,(300-x)=\,(300-x)\,\,mmol\]and \[N{{H}_{4}}C{{l}_{formed}}=x\,\,mmol\] \[N{{H}_{4}}OH\] unreacted \[=300-x-x=(300-2x)\,m\,\,mol\] \[pOH=14-9.26=4.74\] \[p{{K}_{b}}=14-9.26=4.74\] \[pOH=p{{K}_{b}}+\log \,\frac{[NH_{4}^{+}]}{[N{{H}_{4}}OH]}\] \[4.74=4.74+\log \,\frac{x}{(300-2x)}\]\[\frac{x}{300-2x}=1\] \[\therefore \] \[x=100\,mL=\] Volume of HCl \[300-x=200\,mL=\] Volume of \[N{{H}_{4}}OH\] Hence, \[\frac{Volume\,of\,N{{H}_{4}}OH}{Volume\,of\,HCl}=\frac{200}{100}\]\[=\frac{2}{1}\] or 2 : 1You need to login to perform this action.
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