Acid-Base Titration Questions~~ OnLe (1 Viewer)

[Xol_2]

This was found in Top Notes Chemistry, HSC Chemistry Reactions and Calculations, Acidic Earth Chemical Monitoring and Management, pg30, if anyone liked to know the source.

Question, now:
4.8g of anhydrous sodium carbonate was dissolved in water and made up to 200mL in a volumetric flask. 30.0mL of this solution was placed in a cleab conical flask and titrated with sulphuric acid. Calculate the concentration of acid if the 15.1mL were required to reach equivalence.

I've looked at the answer, which is 0.449 mol L^-1 but I'm fighting that it's approximately half of that amount. Because they forgot to multiply 2 mol of hydrogen ion in C1V1 = C2V2... Otherwise, I'm severely confused. (@_@)

 

[Menomaths]

The balanced equation is: Na₂CO₃+H₂SO₄ -> H₂CO₃+Na₂SO₄
Why do you have to multiply by 2?

 

[HeroicPandas]

The balanced equation is: Na₂CO₃+H₂SO₄ -> H₂CO₃+Na₂SO₄
Why do you have to multiply by 2?

Please check ur chemical equation

 

[Menomaths]

Lol break that up into H₂O + CO₂ My bad

Na₂CO₃+H₂SO₄ -> H₂O+CO₂+Na₂SO₄

It's pretty hard typing with all these '_'s!

 

[HeroicPandas]

n(Na2SO4) = m/M = 0.045...mol

[Na2SO4] = n/v = 0.045.../0.2 = 0.226...M(in volumetric flask)

30ml of this 200ml solution is taken out, concentration remains constant,

[Na2SO4] = 0.226...M

Since acid and base in 1:1 mole ratio, u can use C1V1 = C2V2

i.e. [Na2SO4] . V(Na2SO4) = [H2SO4] . V(H2SO4)

Sub in appropriate values,

0.226... x 0.03 = 0.0151 x [H2SO4]

[H2SO4] = 0.449M = 0.045M(2 sig fig)