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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
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B
4.9×102 M
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C
1.9×103 M
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D
2.2×105 M
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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



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