Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • The conductivity of weak acid HA of concentration <img alt="\mathrm{ 0.01 \mathrm{~mol} L^{-1}}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7B%200.01%20%5Cmathrm%7B%7Emol%7

The conductivity of weak acid HA of concentration \mathrm{ 0.01 \mathrm{~mol} L^{-1}}  is  3 \times 10^{-4} \mathrm{scm}^{-1}  If  \mathrm{If~ \lambda \mathrm{~m}(\mathrm{H} A)=190 \mathrm{~s}\mathrm{cm}^2 \mathrm{~mol}^{-1}} Calculate the ionization constant  \mathrm{(k\alpha )}  of  \mathrm{(HA)} . 

Option: 1

\mathrm{ 1.14 \times 10^{-1} }


Option: 2

\mathrm{ 2.66 \times 10^{-1} }


Option: 3

\mathrm{ 3.87 \times 10^{-3} }


Option: 4

\mathrm{ 2.96 \times 10^{-4}}


Answers (1)

best_answer

\mathrm{\lambda_m =1000 \times \frac{K}{M} }
\mathrm{ =1000 \times \frac{3 \times 10^{-4}}{0.01} }
\mathrm{ =30 \text { scm }^2 m o l^{-1} }
\mathrm{\alpha =\frac{\Lambda_m}{\Lambda_m^0}-\frac{30}{190}-\frac{3}{19} }

\mathrm{ H A =H^{+}+A \\ }
\mathrm{ K \alpha =0.01\left(\frac{\alpha^2}{1-\alpha}\right) }
\mathrm{ K \alpha =\frac{0.01 \times\left(\frac{3}{19}\right)^2}{\left[1-\left(\frac{3}{19}\right)\right]} \\ }
\mathrm{ K \alpha =2.96 \times 10^{-4} }.
 

Posted by

chirag

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

Similar Questions

Trending Articles/News

anam-khan

JEE Main 2026 Registration LIVE: NTA session 1 form link at jeemain.nta.nic.in; application deadline
anam-khan

JEE 2026: IIT admissions show major shift toward AI, data science, away from core engineering courses
anam-khan

JEE Main 2026: NIT Delhi cut-offs for BTech courses show ever-increasing demands for AI, computer skills

Latest Question