SATHEE: Chemical and Ionic Equilibrium 2 Question 67

Chemical and Ionic Equilibrium 2 Question 67

67. A $40 mL$ solution of a weak base, $B OH$ is titrated with $0.1 N$ $HCl$ solution. The $pH$ of the solution is found to be 10.04 and 9.14 after the addition of $5.0 mL$ and $20.0 mL$ of the acid respectively. Find out the dissociation constant of the base.

$(1991, 6 M)$

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Solution:

Let $40 mL$ of base contain $x mmol$ of $BOH$ .

$\begin{array}{ccc} BOH + HCl & ⟶ BCl + H_{2} O x - 0.5 & 0.5 & When 5 mL acid is added x - 2 & 2.0 & When 20 mL of acid is added \end{array}$

When $pH$ is $10.04, pOH = 3.96$ and when $pH$ is $9.14, pOH$ is 4.86. Therefore,

$\begin{aligned} 3.96 = p K_{b} + \log \frac{0.50}{x - 0.5} & 3.96 = p K_{b} + \log \frac{2.0}{x - 2} \end{aligned}$

Subtracting Eq. (i) from Eq. (ii) gives

$$ \begin{aligned}

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Chemical and Ionic Equilibrium 2 Question 67

67. A 40 mL solution of a weak base, BOH is titrated with 0.1 N HCl solution. The pH of the solution is found to be 10.04 and 9.14 after the addition of 5.0 mL and 20.0 mL of the acid respectively. Find out the dissociation constant of the base.

Solution:

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67. A $40 mL$ solution of a weak base, $B OH$ is titrated with $0.1 N$ $HCl$ solution. The $pH$ of the solution is found to be 10.04 and 9.14 after the addition of $5.0 mL$ and $20.0 mL$ of the acid respectively. Find out the dissociation constant of the base.