Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • <img alt="\mathrm{\mathrm{K}_{\mathrm{a}_{1}}, \mathrm{~K}_{\mathrm{a}_{2}}}" src="/latex-image/?%5Cmathrm%7B%5Cmathrm%7BK%7D_%7B%5Cmathrm%7Ba%7D_%7B1%7D%7D%2C%20%5Cmathrm%7B%7EK%7D_%7B%5Cmathrm%7B

\mathrm{\mathrm{K}_{\mathrm{a}_{1}}, \mathrm{~K}_{\mathrm{a}_{2}}} and \mathrm{\mathrm{K}_{\mathrm{a}_{3}}} are the respective ionization constants for the following reactions \mathrm{\text { (a), (b), }} and \mathrm{\text { (c). }}

(a) \mathrm{\mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4}=\mathrm{H}^{+}+\mathrm{HC}_{2} \mathrm{O}_{4}^{-}}

(b) \mathrm{HC}_{2} \mathrm{O}_{4}^{-}=\mathrm{H}^{+}+\mathrm{HC}_{2} \mathrm{O}_{4}^{2-}

(c) \mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4}=2 \mathrm{H}^{+}+\mathrm{C}_{2} \mathrm{O}_{4}^{2-}

 The relationship between \mathrm{K}_{\mathrm{a}_{1}}, \mathrm{~K}_{\mathrm{a}_{2}} and \mathrm{K}_{\mathrm{a}_{3}} is given as 

Option: 1

\mathrm{K}_{\mathrm{a}_{3}}=\mathrm{K}_{\mathrm{a}_{1}}+\mathrm{K}_{\mathrm{a}_{2}}


Option: 2

\mathrm{K}_{\mathrm{a}_{3}}=\mathrm{K}_{\mathrm{a}_{1}}-\mathrm{K}_{\mathrm{a}_{2}}


Option: 3

\mathrm{K}_{\mathrm{a}_{3}}=\mathrm{K}_{\mathrm{a}_{1}} / \mathrm{K}_{\mathrm{a}_{2}}


Option: 4

\mathrm{K}_{\mathrm{a}_{3}}=\mathrm{K}_{\mathrm{a}_{1}} \times \mathrm{K}_{\mathrm{a}_{2}}


Answers (1)

best_answer

\mathrm{\text { (a) } \mathrm{H}_2 \mathrm{C}_2 \mathrm{O}_4 \rightleftharpoons \mathrm{H}^{+}+\mathrm{HC}_2 \mathrm{O}_4^{-} \quad: \mathrm{K}_{\mathrm{a}_1}=\frac{\left[\mathrm{H}^{+}\right]\left[\mathrm{HC}_2 \mathrm{O}_4^{-}\right]}{\left[\mathrm{H}_2 \mathrm{C}_2 \mathrm{O}_4\right]}}

\mathrm{\text { (b) } \mathrm{HC}_2 \mathrm{O}_4^{-} \rightleftharpoons \mathrm{H}^{+}+\mathrm{C}_2 \mathrm{O}_4^{2-} \quad: \mathrm{Ka}_2=\frac{[H^{+}]\left[\mathrm{C}_2 \mathrm{O}_4^{2-]}\right]}{\left[\mathrm{HC}_2 \mathrm{O}_4^{-}\right]}}

\mathrm{\text { (C) } \mathrm{H}_2 \mathrm{C}_2 \mathrm{O}_4 \rightleftharpoons 2 \mathrm{H}^{+}+\mathrm{C}_2 \mathrm{O}_4^{2-}} : \mathrm{Ka_{3}=\frac{\left [ \mathrm{H^{+}} \right ]^{2}\left [ \mathrm{C_{2}O_{4}^{2-}} \right ]}{\left [ \mathrm{H_{2}C_{2}O_{4}} \right ]}}

\mathrm{c=a+b \text {, so, } k_{a_3}= k_{a_1} \times k a_2}

Hence, the correct answer is Option (4).

Posted by

Devendra Khairwa

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

AESL expands Aakash Digital 2.0 to offer AI-powered coaching for NEET, JEE, Olympiads
anam-khan

Bihar Board extends BSEB Super 50 free coaching application last date to March 26
anam-khan

JEE Main City Intimation Slip 2025 Live: NTA session 2 exam city slips soon at jeemain.nta.nic.in

Latest Question