1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

# For a 0.01 M solution of carbonic acid, the concentration of CO2−3 would be: Given: pH of the solution is 4.18. Ka1 for H2CO3=4.45×10−7 Ka2 for H2CO3=4.69×10−11 Take 10−4.18=6.6×10−5

A
4.8×1011 M
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
B
6.73×105 M
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
4.5×107 M
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
8.8×1011 M
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

## The correct option is A 4.8×10−11 MGiven, pH=4.18 ⇒−log[H+]=4.18 [H+]=6.6×10−5 Now, H2CO3⇌H++HCO−3 Ka1=[H+][HCO−3][H2CO3] ⇒4.45×10−7=(6.6×10−5)[HCO−3]0.01 ⇒[HCO−3]=6.74×10−5 M Again, HCO−3⇋H++CO2−3 ⇒Ka2=[H+][CO2−3][HCO−3] ⇒[CO2−3]=Ka2[HCO−3][H+] ⇒[CO2−3]=4.79×10−11

Suggest Corrections
32
Join BYJU'S Learning Program
Related Videos
pH and pOH
CHEMISTRY
Watch in App
Explore more
Join BYJU'S Learning Program