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1.2 g mixture of Na_(2)CO_(3) and K_(2)CO_(3) was dissolved in water to form 100 cm^(3) of the solution. 20 cm^(3) of this solution required 40 cm6(3) of 0.1 N HCl solution for neutralisation. Calculate the percentage composition of the mixture.

Answer» <html><body><p></p>Solution :Step I. Calculation of <a href="https://interviewquestions.tuteehub.com/tag/total-711110" style="font-weight:bold;" target="_blank" title="Click to know more about TOTAL">TOTAL</a> no of gram equivalents in the mixture <br/> Let the mass of `Na_(2)CO_(3)` in the mixture = x g <br/> `:.` Mass of `K_(2)CO_(3)` in the mixture `= (1.2 - x)g` <br/> Equivalent mass of `Na_(2)CO_(3)=("<a href="https://interviewquestions.tuteehub.com/tag/molecular-562994" style="font-weight:bold;" target="_blank" title="Click to know more about MOLECULAR">MOLECULAR</a> mass")/(2)=(106)/(2)=53 `<br/> Equivalent mass of `K_(2)CO_(3)=("Molecular mass")/(2)=(138)/(3)=69` <br/> No. of gram equivalents of `Na_(2)CO_(3)=(x)/(53)` <br/> No. of gram equivalent of `K_(2)CO_(3)=((1.2 - x))/(69)` <br/> Total no. of gram equivalent of `Na_(2)CO_(3)` and `K_(2)CO_(3) = (x)/(53) +((1.2 -x))/(69)` <br/> Step II. Calculation of the no. of gram equivalents by titration <br/> No. of gram equivalents in `40 cm^(3)` of 0.1 HCl `= (0.1)/(1000) xx 40 = 4 xx 10^(-3)` <br/> `20 cm^(3)` of the mixture solution <a href="https://interviewquestions.tuteehub.com/tag/neutralise-1114429" style="font-weight:bold;" target="_blank" title="Click to know more about NEUTRALISE">NEUTRALISE</a> `HCl = 4 xx 10^(-3) g eq`. <br/> `100 cm^(3)` of the mixture solution neutralise `HCl = (4 xx 10^(-3))/(20) xx 100 = 2 xx 10^(-2) g aq.` <br/> Since acids and bases react in proportions of their gram equivalents <br/> `:.` No. of gram equivalents of `Na_(2)CO_(3) and K_(2)CO_(3)=2 xx 10^(-2)` <br/> Step III. Calculation of percentage composition of the mixture <br/> Equating eqns. (i) and (ii), `(x)/(53)+((1.2-x))/(69)=2xx10^(-2)=0.02` <br/> `69x+63.6-53x=0.02xx53xx69=73.14` <br/> `16x=73.14-63.6=98.54` <br/> `:. x =(9.54)/(16)=0.596g` <br/> % of `Na_(2)CO_(3)` in the mixture `= (0.596)/(1.2) xx 100 = 49.67` <br/> % of `K_(2)CO_(3)` in the mixture `= 100-49.67 = 50.33`.</body></html>


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