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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 1.9×10−3 MSolution 1 having strong acid HA: pH=4−log[H+]=4[H+]=10−4C1=10−4 M Solution B having strong acid HB: Let concentration be C2 Final solution obtained after mixing pH=3−log[H+]=3[H+]=10−3 M Concentration (Cf)=10−3 M 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=1.9×10−3 M

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