wiz-icon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

Consider a strong acid HA having a pH=4 mixed in an equal volume with another solution of strong acid HB to give a resultant pH of 3. Then the H+ concentration of solution containing strong acid HB is :

A
3.5×103 M
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
4.9×102 M
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
1.9×103 M
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
D
2.2×105 M
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is C 1.9×103 M
Solution 1 having strong acid HA:
pH=4log[H+]=4[H+]=104C1=104 M
Solution B having strong acid HB:
Let concentration be C2
Final solution obtained after mixing
pH=3log[H+]=3[H+]=103 M

Concentration (Cf)=103 M
The volume of two solutions are equal.
So,
V1=V2=VVf=V1+V2=2VCfVf=C1V1+C2V2103×2V=(104×V)+(C2×V)C2=(2×103)(104)C2=1.9×103 M



flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
Join BYJU'S Learning Program
CrossIcon