Question

A solution contains Na2CO3 and NaHCO3. 10ml of the solution requires 2.5ml of 0.1M H2SO4 for neutralisation using phenolphthalein as an indicator. Methyl orange is added when a further 2.5ml of 0.2M H2SO4 was required. Calculate amount of Na2CO3 and NaHCO3 in 1L of solution.​

Answer 1

Answer:

Explanation:

With phenolpthalene, the reaction occurring is

Na2CO3 + H+ → NaHCO3 + Na

To neutralise whole of Na2CO3, volume of H2SO4 will be twice of the given volume

N1 = (2×2.5)(2×0.1)10=0.1 N

Mass of Na2CO3 in 1L solution is =0.1×53=5.3 g

Volume of sulphuric acid solution used in neutralising

NaHCO3==2.5 ml

Hence, normality of NaHCO3 solution,

N2=2.5×(2×0.2)10=0.1 NMass of NaHCO3 = 0.1×84=8.4 g

Thank you

Answer 2

Holaaa Mate

Answer:

With phenolpthalene, the reaction occurring isNa2CO3 + H+ → NaHCO3 + Na+To neutralise whole of Na2CO3, volume of H2SO4 will be twice of the given volumeN1 = (2×2.5)(2×0.1)10=0.1 NMass of Na2CO3 in 1L solution is =0.1×53=5.3 gVolume of sulphuric acid solution used in neutralising NaHCO3==2.5 mlHence, normality of NaHCO3 solution, N2=2.5×(2×0.2)10=0.1 NMass of NaHCO3 = 0.1×84=8.4 g