SATHEE: Chemical and Ionic Equilibrium

Chemical and Ionic Equilibrium - Result Question 123

####67. A $40 m L$ solution of a weak base, $B O H$ is titrated with $0.1 N$ $H C l$ solution. The $p H$ of the solution is found to be 10.04 and 9.14 after the addition of $5.0 m L$ and $20.0 m L$ of the acid respectively. Find out the dissociation constant of the base.

$(1991, 6 M)$

Show Answer

Solution:

Let $40 m L$ of base contain $x m m o l$ of $B O H$ .

$\begin{array}{ccc} B O H + H C l & ⟶ B C l + H_{2} O x - 0.5 & 0.5 & When 5 m L acid is added x - 2 & 2.0 & When 20 m L of acid is added \end{array}$

When $p H$ is $10.04, p O H = 3.96$ and when $p H$ is $9.14, p O H$ 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}

Play Store

App Store

Ask SATHEE

Welcome to SATHEE !
Select from 'Menu' to explore our services, or ask SATHEE to get started. Let's embark on this journey of growth together! 🌐📚🚀🎓

I'm relatively new and can sometimes make mistakes.
If you notice any error, such as an incorrect solution, please use the thumbs down icon to aid my learning.
To begin your journey now, click on "I understand".

Chemical and Ionic Equilibrium - Result Question 123

Solution:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

Menu