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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) |
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Answer» `2.7xx10^(-2)M` 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")):}` 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). `"Concentration of salt" = ("Millimoles of base")/("TOTAL volume")` Millimoles of base `=M.V_(mL)` `=((2)/(5))(2.5)` Total volume `=V_("acid")+V_("base")` `V_("acid")=(M_("base")V_("base"))/(M_("acid"))` `=((2//5)(2.5))/((2//15))=7.5 mL` `:. C_("salt")=((2//5)(2.5))/((7.5+2.5))=(1)/(10)=0.1` For cationic hydrolysis, `K_(h)=(Ch^(2))/(1-h)=(K_(w))/(K_(b))=(10^(-14))/(10^(-12))=10^(-2)` We have `C=0.1`, `K_(w)=1xx10^(-14)`, `K_(b)=1xx10^(-12)` This gives `h=0.27` (since `K_(h)` is not neglected relative to 1) `C_(H^(+))=Ch` `=(0.1)(0.27)` `=2.7xx10^(-2)M` If we neglact h relative 1, the answer will be `3.2xx10^(-2)`M. |
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