InterviewSolution
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InterviewSolution
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Following solutions were prepared by mixing different volumes of NaOH and HCl of different concentrations: a. 60 mL `(M)/(10) HCl + 40 mL (M)/(10)` NaOH b. 55 mL `(M)/(10) HCl + 45 mL (M)/(10)` NaOH c.75 mL `(M)/(5) HCl + 25 mL (M)/(5)` NaOH d. 100 mL `(M)/(10) HCl + 100 mL (M)/(10)` NaOH pH of which one of them will be equal to 1 ?A. IVB. IC. IID. III |
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Answer» Correct Answer - D `75 mL(M)/(5) HCl+25mL (M)/(5)NaOH` Milliequivalent of HCl `=75 mL " of" (M)/(5) HCl=(1)/(5)xx75=15` Milliequivalent of NaOH `=25 mL " of" (M)/(5) NaOH` `=(1)/(5)xx25=5` `therefore` Milliequivalent of HCl left unused =15-5=10 Volume of solution =100 mL `therefore` Molarity of `[H^(+)]` in the resulting mixture `=(10)/(100)=(1)/(10)` `therefore " " pH="log" (1)/([H^(+)])` =log (10)=1 |
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