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6.2. Equilibrium - II (icon Equilibrium) — Chemistry STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceChemistry6.2. Equilibrium - II (icon Equilibrium)1 Mark
MCQ
A solution of $0.1 M$ weak base $( B )$ is titrated with $0.1 M$ of a strong acid ($HA$). The variation of $pH$ of the solution with the volume of $HA$ added is shown in the figure below. What is the $p K_{ b }$ of the base? The neutralization reaction is given by $B + HA \rightarrow BH ^{+}+ A ^{-}$.
  • A
    $2.60$
  • ✓
    $2.80$
  • C
    $2.85$
  • D
    $2.90$
✓

Answer

Correct option: B.
$2.80$
b
$B + HA \longrightarrow BH ^{+}+ A ^{-}$

$0.1 M , V ml$

$0.1 V m mol \quad 0.1 V m mol \quad 0.1 V 0.1 V$

${\left[ BH ^{+}\right]=\frac{0.1 V }{2 V }=0.05 M }$

$pH$ at eq. $pt =6$ to 6.28

$pH =7-\frac{1}{2}\left[ pK _{ b }+\log 0.05\right]$

So $pK _{ b }=2.30-2.80$

Possible

Solution-$2$

at $V =6 ml \quad rxn$ is complete

So $V =3 ml$ is half of eq. pt at which $\quad pH =11$

$pOH =(14-11)= pK _{ b }+\log 1$

$pK _{ b }=3$

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