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A sample of mixed alkalis containing NaOH and Na_(2)CO_(3) is titrated in the following two schemes : (i) 10 ml of above mixture requires 8 ml of 0.1 N HCl by using phenolphthalein. (ii) 10 ml of above mixture requires 10 ml of 0.1 N HCl by using methyl orange. Calculate the ratio of the weight of NaOH and Na_(2)CO_(3) in the sample mixture.

Answer» <html><body><p></p>Solution :Basic principle involved is as follows : <br/> Acid used with <a href="https://interviewquestions.tuteehub.com/tag/phenolphthalein-599562" style="font-weight:bold;" target="_blank" title="Click to know more about PHENOLPHTHALEIN">PHENOLPHTHALEIN</a> as <a href="https://interviewquestions.tuteehub.com/tag/indicator-1041513" style="font-weight:bold;" target="_blank" title="Click to know more about INDICATOR">INDICATOR</a> = Complete neutralization ofNaOH `+ (1)/(2)` neutralization of `Na_(2)CO_(3)` <br/> Acid used with methyl orange as indicator= Complete neutralization of NaOH + complete neutralization of `Na_(2)CO_(3)` <br/> <a href="https://interviewquestions.tuteehub.com/tag/meq-1093952" style="font-weight:bold;" target="_blank" title="Click to know more about MEQ">MEQ</a> of HCl used for 10 ml of mixture using phenolphthalein `= 8 <a href="https://interviewquestions.tuteehub.com/tag/xx-747671" style="font-weight:bold;" target="_blank" title="Click to know more about XX">XX</a> 0.1 = 0.8` <br/> `:. ` meq of`NaOH + (1)/(2) M "eq of"Na_(2)CO_(3) = 0.8`...(i)<br/> meq of HCl usedfor 10 ml of mixture using methly orange `= 10 xx 0.1 = 1 ` <br/> `:.` meq of NaOH + M eq of `Na_(2)CO_(3)=1` <br/> From eqns (i) and (<a href="https://interviewquestions.tuteehub.com/tag/ii-1036832" style="font-weight:bold;" target="_blank" title="Click to know more about II">II</a>), meq of `Na_(2)CO_(3) = (1-0.8)xx2=0.4 ,` meq of `NaOH = 1 - 0.4 = 0.6` <br/> Eq. wt of `Na_(2)CO_(3) = 106//2=53`, Eq. wt of NaOH = 40 <br/> `:. ` 0.4 meq of `Na_(2)CO_(3) = (53)/(1000) xx0.4 = 0.0212 g ` <br/> 0.6meq of `NaOH = (40)/(1000) xx 0.6 = 0.024 g` <br/> Ratio of weight of NaOH and `Na_(2)CO_(3)=0.024//0.0212 = 1.132`</body></html>


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