Q.

What % of the carbon in the H2CO3 - HCO3- buffer should be in the form of HCO3- so as to have a neutral solution ? Ka= 4 × 10-7

see full answer

Start JEE / NEET / Foundation preparation at rupees 99/day !!

21% of IItians & 23% of AIIMS delhi doctors are from Sri Chaitanya institute !!
An Intiative by Sri Chaitanya

a

20%

b

40%

c

60%

d

80%

answer is D.

(Unlock A.I Detailed Solution for FREE)

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

Detailed Solution

pH=pKa + log HCO3-H2CO3 ⇒  7=7-log 4 + log  HCO3-H2CO3⇒HCO3-H2CO3=4 %  of carbon in the form of HCO3- = HCO3-HCO3-H2CO3 × 100 = 41+4 × 100=80%
Watch 3-min video & get full concept clarity
Related Blogs
Get ₹ 1 crore* worth scholarship for JEE / NEET / Foundation
Result Producing online coaching for JEE/NEET of south India
Largest All India Test Series for JEE/NEET
Best Books for IIT JEE
Best Books for NEET
How to Join IIT after 10th
score_test_img

Get Expert Academic Guidance – Connect with a Counselor Today!

Related Blogs
Get ₹ 1 crore* worth scholarship for JEE / NEET / Foundation
Result Producing online coaching for JEE/NEET of south India
Largest All India Test Series for JEE/NEET
Best Books for IIT JEE
Best Books for NEET
How to Join IIT after 10th
whats app icon
What % of the carbon in the H2CO3 - HCO3- buffer should be in the form of HCO3- so as to have a neutral solution ? Ka= 4 × 10-7