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Characteristics of Equilibrium Constant

A weak base ...

Question

A weak base $(50.0 m L)$ was titrated with $0.1 M H C l$ . The $p H$ of the solution after the addition of $10.0 m L$ and $25.0 m L$ were found to be $9.84$ and $9.24$ , respectively. Calculate $K_{b}$ of the base and $p H$ at the equivalence point.

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Solution

$(I)$
$B O H + H C l ⟶ B C l + H_{2} O a 0.1 \times 1000 a - 1011$
$p H = 9.8 M$ $p O H = 4.16$
$p O H = - l o g K_{b} + l o g [\frac{B C l}{B O H}]$
$4.15 = - l o g K_{b} + l o g [\frac{1}{a - 1}]$
$(I I)$
$B O H + H C l ⟶ B C l + H_{2} O a 0.1 \times 2500 a - 2.5 0 2.5 2.5$

$p H = 9.24$ $p O H = 4.76$
$4.70 = - l o g K_{b} + l o g [\frac{.5}{a - 0.25}]$

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