1.

50 ml of 0.05 M HNO_3 is added to 50 ml of 0.025 M KOH. Calculate the pH of the resultant solution.

Answer»

SOLUTION :Number of MOLES of `HNO_3=0.05 times 50 times 10^-3=2.5 times 10^-3`
Number of moles of KOH=`0.025 times 50 times 10^-3=1.25 times 10^-3`
Number of moles of `HNO_3` after mixing =`2.5 times 10^-3-1.5 times 10^-3=1.25 times 10^-3`
`therefore " Concentration of " HNO_3= ("Number of moles of" HNO_3)/( "Volume in litre")`
After mixing total volume=`1.25 times 10^-2moles L^-1`
`therefore[H^+]=(1.25 times10^-3 MOL es)/(100 times10^-3L)=1.25 times10^-2mol es L^-1`
`pH=-log[H^+]`
`pH=-log(1.25 times10^-2)=2-0.0969`
`=1.9031`


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