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200 " mL of " a solution of a mixture of...

200 " mL of " a solution of a mixture of NaOH and `Na_2CO_3` was first titrated with 0.1 M HCl using phenolphthalein indicator. 17.5 " mL of " HCl was required for the same HCl was again required for next end point. Find the amount of NaOH and `Na_2CO_3` in the mixture.

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Let , `W_(1)` gm NaOH and `W_(2) gm Na_(2)CO_(3)` was present in mixture . At phenophthaiein end point ,
`(W_(1)/40 + 1/2 xx W_(2)/53 ) = 1/10 xx 17.5 xx 10^(-3)`
At second end point following reaction takes place
Eq. of `NaHCO_(3) = Eq`. of HCl used (in second titration )
` = 1/2 " Eq. of " Na_(2)CO_(3)`
` 1/2 xx W_(2)/53 = 25 xx 1/10 xx 10^(-3) `
`W_(2) = 0.0265` gm
Putting the value of `W_(2) ` in Eq. (1) , we get
`W_(1) = 0.06` gm
Percentage of NaOH = `(0.06)/(0.2) xx 100 = 30 %`
Percentage of `NaCO_(3) = (0.0265)/(0.2) xx 100 = 13.25 %`
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