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30 cc of M/3 HCl, 20 cc of M/2 HNO_(3) and 40 cc of M/4 NaOH solutions are mixed and the volume wasmade up of 1 dm^(3) . The pH of the resulting solution is

Answer»

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SOLUTION :Total milliequivalents of `H^(+)`
`=20xx1/2+30xx1/3=20`
Total milliequivalents of `OH^(-)=40xx1/4=10`
Thus, milliequivalents of `OH^(-)=40xx1/4=10`
Thus, milliequivalents of `H^(+)`left `=20-10=10`
Total VOLUME of solution `=1dm^(3)=1000` ml
`:.[H^(+)]=10/1000=10^(-2)`
`impliespH=-LOG[H^(+)]=-log[10^(-2)]=2`


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