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InterviewSolution
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A mixutre solution of KOH and `Na_2CO_3` requires 15 " mL of " `(N)/(20)` HCl when titrated with phenolphthalein as indicator.But the same amoound of the solutions when titrated with methyl orange as indicator requires 25 " mL of " the same acid. Calculate the amount of KOH and `Na_2CO_3` present in the solution. |
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Answer» (a). With methyl orange as indicator: Full titre value of KOH (let x mL)`+` full titre value of `Na_2CO_3` (let y mL)`=25mL` `x+y=25` .(i) (b). With phenolphthalein as indicator: Full titre value of KOH(x)`+(1)/(2)` titre value of `Na_2CO_3` `((y)/(2))=15mL` `x+(y)/(2)=15` ...(ii) Solving (i) and (ii) we get `x=5mL, y=20mL` Full titre value of `Na_2CO_3=20mL` m" Eq of "`Na_2CO_3=mEq ` of HCl `=20xx(N)/(20)mEq` of `Na_2CO_3` `=1xx10^(-3)xx53g of Na_2CO_3` `=0.053g` Volume of KOH`=25-20=5mEq` of `KOH` `=5xx10^(-3)xx56g` of `KOH` `=0.014` g of `KOH` |
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