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
3.5×10−3M
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
4.9×10−2M
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
1.9×10−3M
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
D
2.2×10−5M
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution
The correct option is C1.9×10−3M Solution 1 having strong acid HA: pH=4−log[H+]=4[H+]=10−4C1=10−4M
Solution B having strong acid HB:
Let concentration be C2
Final solution obtained after mixing pH=3−log[H+]=3[H+]=10−3M
Concentration (Cf)=10−3M
The volume of two solutions are equal.
So, V1=V2=VVf=V1+V2=2VCfVf=C1V1+C2V210−3×2V=(10−4×V)+(C2×V)C2=(2×10−3)−(10−4)C2=1.9×10−3M