A 50 ml sample of NaHCO3 and Na2CO3 is titrated with 12ml, 0.2M H2SO4

1.	A 50 ml sample of NaHCO3 and Na2CO3 is titrated with 12ml, 0.2M H2SO4 using methyl Orange to reach end Point.
Answer» In case of phenolphathalin indicator - only Na₂CO₃ react with H₂SO₄ → 2NaHCO₃ + Na₂SO₄ So number of M moles of Na₂CO₃ = Number of M moles of H⁺ ions = 2 x (0.2 x 4ml) ⇒ 1.6 M mole In case of method orange - both NaHCO₃ and Na₂Ca₃ will react. 2NaHCO₃ + H₂SO₄ → Na₂SO₄ + H₂O + CO₂--- (i) Na₂CO₃ + H₂SO₄ → Na₂SO₄ + H₂O + CO_2----(ii) Total M mole of H⁺_ion = 2(12ml x 0.2 M) = 2(2.4 M mole) = 4.8 M mole from eqn ---(ii) \(\because\) 1 mole of Na₂CO₃ required to neutralize = 2 mole of H⁺_ions \(\therefore\)1.6 M mole Na₂CO₃ required to neutralize = 3.2 M mole of H⁺_ions from eqn---(i) we can see that----- 1 mole of NaHCO₃ required to neutralize = 1 mole of H⁺_ions So, Number of M moles of NaHCO₃ = (4.8 - 3.2) = 1.6 M mole Therefore the ration of M mole of NaHCO₃ to Na₂CO₃ ⇒ 1.6/1.6 = 1 : 1

A 50 ml sample of NaHCO3 and Na2CO3 is titrated with 12ml, 0.2M H2SO4 using methyl Orange to reach end Point.

Answer»

In case of phenolphathalin indicator - only

Na₂CO₃ react with H₂SO₄ → 2NaHCO₃ + Na₂SO₄

So number of M moles of Na₂CO₃ = Number of M moles of H⁺ ions

= 2 x (0.2 x 4ml)

⇒ 1.6 M mole

In case of method orange - both NaHCO₃ and Na₂Ca₃ will react.

2NaHCO₃ + H₂SO₄ → Na₂SO₄ + H₂O + CO₂--- (i)

Na₂CO₃ + H₂SO₄ → Na₂SO₄ + H₂O + CO_2----(ii)

Total M mole of H⁺_ion = 2(12ml x 0.2 M)

= 2(2.4 M mole)

= 4.8 M mole

from eqn ---(ii)

\(\because\) 1 mole of Na₂CO₃ required to neutralize = 2 mole of H⁺_ions

\(\therefore\)1.6 M mole Na₂CO₃ required to neutralize = 3.2 M mole of H⁺_ions

from eqn---(i) we can see that-----

1 mole of NaHCO₃ required to neutralize = 1 mole of H⁺_ions

So, Number of M moles of NaHCO₃ = (4.8 - 3.2) = 1.6 M mole

Therefore the ration of M mole of NaHCO₃ to Na₂CO₃

⇒ 1.6/1.6 = 1 : 1