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2.5 mL of 2//5 M weak monoacidic base (K_(b)=1xx10^(-12) at 25^(@)C) is titrated with 2//15 M HCI in water at 25^(@)C. The concentration of H^(+) at equivalence point is (K_(w)=1xx10^(-14) at 25^(@)C)

Answer» <html><body><p>`<a href="https://interviewquestions.tuteehub.com/tag/2-283658" style="font-weight:bold;" target="_blank" title="Click to know more about 2">2</a>.7xx10^(-2)M`<br/>`3.2xx10^(-2)M`<br/>`3.2xx10^(-7)M`<br/>`2.7xx10^(-13)M`</p>Solution :Weak monoacidic base such as `NH_(4)OH` (aqueous `NH_(3)`) is neutralized by HCI as follows: `NH_(4)OH(aq.)+HCl(aq.)rarrNH_(4)<a href="https://interviewquestions.tuteehub.com/tag/cl-408888" style="font-weight:bold;" target="_blank" title="Click to know more about CL">CL</a>(aq.)+H_(2)O(l)` <br/> At the equivalence point, all `NH_(4)OH` gets converted into `NH_(4)Cl` (salt) which undergoes cationic hydrolysis as follows: `{:(,NH_(4)^(+)(aq.)+H_(2)O(l)hArrNH_(4)OH(aq.)+H^(+)(aq.)),("Initial (M)","C00"),("Change (M)", "-Ch+Ch+Ch"),("Equilibrium (M)",bar("C(1-h)ChCh")):}` <br/> Thus, the concentration of `H^(+)` ions is due to the hydrolysis of the resulting salt `(NH_(4)Cl)`. To calculate is value, we need the concentration of salt (C ) and the degree of hydrolysis of salt (h). <br/> `"Concentration of salt" = ("Millimoles of base")/("<a href="https://interviewquestions.tuteehub.com/tag/total-711110" style="font-weight:bold;" target="_blank" title="Click to know more about TOTAL">TOTAL</a> volume")` <br/> Millimoles of base `=M.V_(mL)` <br/> `=((2)/(5))(2.5)` <br/> Total volume `=V_("acid")+V_("base")` <br/> `V_("acid")=(M_("base")V_("base"))/(M_("acid"))` <br/> `=((2//5)(2.5))/((2//15))=7.5 mL` <br/> `:. C_("salt")=((2//5)(2.5))/((7.5+2.5))=(1)/(10)=0.1` <br/> For cationic hydrolysis, <br/> `K_(h)=(Ch^(2))/(1-h)=(K_(w))/(K_(b))=(10^(-<a href="https://interviewquestions.tuteehub.com/tag/14-272882" style="font-weight:bold;" target="_blank" title="Click to know more about 14">14</a>))/(10^(-12))=10^(-2)` <br/> We have <br/> `C=0.1`, `K_(w)=1xx10^(-14)`, `K_(b)=1xx10^(-12)` <br/> This gives `h=0.27` (since `K_(h)` is not neglected relative to 1) <br/> `C_(H^(+))=Ch` <br/> `=(0.1)(0.27)` <br/> `=2.7xx10^(-2)M` <br/> If we neglact h relative 1, the answer will be `3.2xx10^(-2)`M.</body></html>


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