2.5mL of (2)/(5) M weak monoacidic base (K_(b)=1xx10^(-12) "at" 25^(@)C) is titrated with (2)/(15) M HCl in water at 25^(@) C. The concentration of H^(+) at equivalence point is (K_(w) = 1 xx 10^(-14)"at"25^(@)C)

2.5mL of (2)/(5) M weak monoacidic base (K_(b)=1xx10^(-12) "at" 25^(@)

1.	2.5mL of (2)/(5) M weak monoacidic base (K_(b)=1xx10^(-12) "at" 25^(@)C) is titrated with (2)/(15) M HCl in water at 25^(@) C. The concentration of H^(+) at equivalence point is (K_(w) = 1 xx 10^(-14)"at"25^(@)C)
Answer» `3.7xx10^(-13)M` `3.2xx10^(-7) M` `3.2xx10^(-2)M` `2.7xx10^(-2)M` Solution :`{:(N_(1)V_(1),=,N_(2)V_(2)),("(Base)",,"(ACID)"):}` `2.5xx(2)/(5)=(2)/(15)=(2)/(15)xxV_(2) "or" V_(2)=(15)/(2) ML = 7.5 mL ` `BOH+HClrarrBCl+H_(2)O` 2.5 mL of 2 M base contain base `=2.5xx(2)/(5) = 1` mmol `:. ` Salt BCl formed = 1 mmol Volume of solution = 2.5 mL + 7.5 mL = 10 mL `:.` Conc. of salt [BCl] in the solution `=(1)/(10)M=0.1M` For salt of weak base and STRONG acid `[H^(+)]=sqrt((K_(w)C)/(K_(b)))=sqrt((10^(-14)xx0.1)/(10^(-12)))=3.2xx10^(-2)M`