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# Acid & bases help CHEM4 AQA Watch

1. How do you work out question 6e?

Volume of HCl(aq) added for first end-point 15
Volume of HCl(aq) added for second end-point 45
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2. 4.3 exam questions (1).rtf (35.6 KB, 111 views)
3. (Original post by Smcdonald98)
How do you work out question 6e?

Volume of HCl(aq) added for first end-point 15
Volume of HCl(aq) added for second end-point 45
The first inflexion is due to the reaction:

Na2CO3 + HCl --> NaHCO3 + NaCl

The addition of more sodium hydrogencarbonate is not going to affect this, as it only shows the reaction of sodium carbonate. The inflexion remains at 15ml

As the first inflexion is at 15ml adding an equal number of moles of sodium hyrogencarbonate will add an extra 15ml to the second inflexion = 45ml
4. (Original post by charco)
The first inflexion is due to the reaction:

Na2CO3 + HCl --> NaHCO3 + NaCl

The addition of more sodium hydrogencarbonate is not going to affect this, as it only shows the reaction of sodium carbonate. The inflexion remains at 15ml

As the first inflexion is at 15ml adding an equal number of moles of sodium hyrogencarbonate will add an extra 15ml to the second inflexion = 45ml

A titration curve is plotted showing the change in pH as a 0.0450 mol dm–3 solution of sodium hydroxide is added to 25.0 cm3 of a solution of ethanedioic acid, H2C2O4 The titration curve obtained has two equivalence points (end points)
(ii) When the second equivalence point is reached, a total of 41.6 cm3 of 0.0450 mol dm–3 sodium hydroxide has been added. Calculate the concentration of the ethanedioic acid solution.

In this question why is the moles of H2C2O4 equal to half moles of OH–? Thanks
5. (Original post by Smcdonald98)
A titration curve is plotted showing the change in pH as a 0.0450 mol dm–3 solution of sodium hydroxide is added to 25.0 cm3 of a solution of ethanedioic acid, H2C2O4 The titration curve obtained has two equivalence points (end points)
(ii) When the second equivalence point is reached, a total of 41.6 cm3 of 0.0450 mol dm–3 sodium hydroxide has been added. Calculate the concentration of the ethanedioic acid solution.

In this question why is the moles of H2C2O4 equal to half moles of OH–? Thanks
Have you tried writing out the equation for the reaction?

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