titration question?

Question

25.0 cm (cube) of a solution containing sodium hydroxide and sodium carbonate required 21.50 cm(cube) of 1mol/dm(cube) hcl with phenolphtalein as indicator and 27.85cm(cube) of the same acid with methyl orange as indicator. caculate the concentration of each compound in g/dm(cube) of the solution.

plz give the steps and explanation

Answer 1

Sodium hydroxide / sodium carbonate solution
V= 25.0 mL
Composition= ???

Titration 1
Indicator: Phenolphthalein
V of HCl = 21.5 mL
CM = 1.0 M

Titration 2
Indicator: Methyl Orange
V of HCl = 27.85 mL
CM = 1.0 M

During titration of the sodium hydroxide and sodium carbonate solution the following reactions take place:

1- HCl + NaOH -----> NaCl + water
2- HCl + sodium carbonate -----> NaCl + sodium hydrogen carbonate
3- HCl + sodium hydrogen carbonate ----->NaCl + carbon dioxide + water


Indicator / pH range over which the indicator changes color
---------------------------------------------------------------------------
Methyl Orange / 3.2 - 4.5
Phenolphthalein / 8.2 - 10
---------------------------------------------------------------------------

The "end point" of titration usually occurs when one drop of the titrant is added beyond the "equivalence point". This happens when 1 drop of HCl is added turning the medium slightly acid.

The color indicator that changes color in a slightly acidic medium is methyl orange (not phenolphthalein). It indicates the end of titration (i.e when all the bases in the solution are neutralized).

- Methyl orange changes color (from pink to colorless) when reactions 1, 2 and 3 are complete.
- Phenolphthalein changes color from (yellow to pink) when reactions 1, and 2 are complete.

The difference in the volume of the acid titrant between titration 1 and titration 2 is nothing but the volume of the acid required for reaction 3.

Volume of the acid titrant needed to neutralize the sodium hydrogen carbonate (rxn 3)
---------------------------------------------------------------------------------------------------------
27.85 mL – 21.50 mL = 6.35 ml

Volume of the acid titrant needed to neutralize the sodium carbonate
--------------------------------------------------------------------------------------------
2 x 6.35 ml = 12.7 mL

Volume of the acid titrant needed to neutralize the sodium hydroxide
------------------------------------------------------------------------------------------------------------------------------------------------------------------------
27.85 mL – 12.7 mL = 15.15 ml

===============================================

HCl + NaOH -----> NaCl + water

Moles of NaOH = moles of HCl ( 1:1 ratio)
= (CM x V) acid
= 1.00 mol/L x 0.01515 L
= 0.0152 mol.

Concentration of NaOH(mol/L) = mol. / Volume
= 0.0152 mol. / 0.025 L
= 0.608 M

Concentration of NaOH (g/L) = Concentration of NaOH(mol/L) x Molar Mass (g/mol)
= 0.608 mol./ L x 40.0 g/mol.
= 24.3 g/L
Note that 1 L = 1 dm(cube)
===============================================

HCl + sodium carbonate -----> NaCl + carbon dioxide + water

Moles of Sodium carbonate = (1/2) x moles of HCl (1:2 ratio)
= (1/2)x(CM x V) acid
= (1/2)x1.00 mol/L x 0.0127 L
= 0.0635 mol.

Concentration of sodium carbonate(mol/L) = mol. / Volume
= 0.0635 mol. / 0.025 L
= 0.254 M

Concentration of sodium carbonate (g/L) = Concentration of sodium carbonate (mol/L) x Molar Mass (g/mol)
= 0.254 mol./ L x 106.0 g/mol.
= 26.9 g/L
Note that 1 L = 1 dm(cube)

Sorry for any typing mistake!
Sam M

Answer 2

Phenolphthalein will detect:
OH- + H+ ---> H2O and (CO3)2- + H+ ----> (HCO3)-

Methyl orange will detect:
OH- + H+ ---> H2O and (CO3)2- + 2H+ ----> CO2 + H2O

Now perhaps you can see the significance in the difference between the two titres - it's how much hydrogencarbonate there was (and carbonate).

Now use the figures to work out how much hydroxide there was.