The overall equation for the reaction between sodium carbonate solution and dilute hydrochloric acid is Na_(2)CO_(3)(aq.)+2HCl(aq.)rarrNaCl(aq)+CO_(2)(g)H_(2)O(l) If you had the two solutions of the same concentration, you would have to use double volume of HCl to reach the equivalence point. Indicators change their colours at the end point of the reaction and hence we are able to know the end points (equivalence points of reactions). 25 ml of Na_(2)CO_(3) solution requires 100 ml of 0.1 M HCl to reach end point with phenolphthaleinindicator. Molarity of HCO_(3^(-)) ions in the resulting solution is

The overall equation for the reaction between sodium carbonate solutio

1.	The overall equation for the reaction between sodium carbonate solution and dilute hydrochloric acid is Na_(2)CO_(3)(aq.)+2HCl(aq.)rarrNaCl(aq)+CO_(2)(g)H_(2)O(l) If you had the two solutions of the same concentration, you would have to use double volume of HCl to reach the equivalence point. Indicators change their colours at the end point of the reaction and hence we are able to know the end points (equivalence points of reactions). 25 ml of Na_(2)CO_(3) solution requires 100 ml of 0.1 M HCl to reach end point with phenolphthaleinindicator. Molarity of HCO_(3^(-)) ions in the resulting solution is
Answer» 0.008 M 0.04M 0.16 M 0.08M Solution :In presence of phenophthalein indicator valence factor of HPH = 1 EQUIVALENTS of HCl = Equivalent of `Na_(2)CO_(3)` `N_(1)V_(1)=N_(2)V_(2)` For HCl `N_(1)=M_(1) ` and `Na_(2)CO_(3)N_(2)=N_(2)M_(1)V_(1)=M_(2)V_(2)` `0.1 xx 100 = N_(2)xx25` `M_(2)=0.4` Reaction is `Na_(2)CO_(3) +HCl rarrNaHCO_(3)+NaCl` in moles of `Na_(2)CO_(3)=MV` `0.4 018xx25xx10^(-3)` =moles of `NaCHO_(3)` molarity of `HCO_(3)^(-)=("Moles of " NaHCO_(3))/("Volume")=(0.4xx25xx10^(-3)xx1000)/(125)=0.08`