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2.5 mL of (2)/(5) M weak monoacidic bas...

`2.5 ` mL 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)`

A

`3.7xx10^(-13)` M

B

`3.2xx10^(-7)` M

C

`3.2xx10^(-2)` M

D

`2.7xx10^(-2)` M

Text Solution

Verified by Experts

The correct Answer is:
C
At equivalence `N_(a)V_(a)=N_(b)V_(b)`
`(2)/(15)xxV_(a)=(2)/(5)xx2.5`
`V_(a)=7.5` mL.
Moles of acid = Moles of base = `2.5xx2//5xx10^(-3)=10^(-3)`
Volume of solution = `2.5+7.5=10` mL
=`10^(-2)L`
`underset(0)BOH+underset(0)HCl hArr underset(10^(-3)) BCl+underset(10^(-3))(H_(2)O` Final concentration
Concentration of salt = `(10^(-3))/(10^(-2))"moles/L"`
`[C]=0.1` M
`pH=7-(1)/(2)(pK_(b)+log C)`
= `7-(1)/(2)(-log 10^(-12)+log(0.1))`
= `1.5`
`[H^(+)]=10^(-ph)=10^(-1.5)=3.2xx10^(-2)`
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