You are Here: Home

# edexcel acid/base equilibria watch

1. some qs on acid and base equilibiria:

1) i don't understand the principle of the "half volume method" to find the Ka . In the book , it states that in involves titrating until half of the molecules of the weak acid have reacted. the concentration of the weak acid molecules and its conjugate base ions are equal" How come the two concentrations are equal?
HA--> H+ + A-
Is it because when half of the total amount of HA is reacted, the remaning HA will dissociate into Hydrogen ion and A-, so A- and HA have the same concentration. but isn't it (the situation) the same when the HA has not been reacted?

2)why the ionic product of water must be 10^-14? and why this value can be applicable to other solution (acid and base)?

3)why for equimolar of acid and base, regardless of their strengths (strong /weak), same volume of solution is used to reach the equivalence point. what i originall thought was that for examplem, in weak acid-strong base reaction, though they are equimolar, but weak acid dissociates less in the aqueous solution, so more volume should be used to get the equivalence point? can someone explain this to me ?

Thank you soooooo much!!!!
2. (Original post by Lamalam)
some qs on acid and base equilibiria:

1) i don't understand the principle of the "half volume method" to find the Ka . In the book , it states that in involves titrating until half of the molecules of the weak acid have reacted. the concentration of the weak acid molecules and its conjugate base ions are equal" How come the two concentrations are equal?
HA--> H+ + A-
Is it because when half of the total amount of HA is reacted, the remaning HA will dissociate into Hydrogen ion and A-, so A- and HA have the same concentration. but isn't it (the situation) the same when the HA has not been reacted?

2)why the ionic product of water must be 10^-14? and why this value can be applicable to other solution (acid and base)?

3)why for equimolar of acid and base, regardless of their strengths (strong /weak), same volume of solution is used to reach the equivalence point. what i originall thought was that for examplem, in weak acid-strong base reaction, though they are equimolar, but weak acid dissociates less in the aqueous solution, so more volume should be used to get the equivalence point? can someone explain this to me ?

Thank you soooooo much!!!!

At the equivalence point all of the acid has reacted and produced salt.

Hence the moles of acid (originally) = moles of salt formed

When half the moles of acid (moles/2) have reacted they have also left behind half the moles of acid (moles/2) AND formed half the moles of salt (moles/2.

Hence moles of acid remaining = moles of salt formed

They are both in the same solution (volume must be the same), so:

[HA] = [A-]

At this point

[H+] = Ka

pH = pKa
3. Actually it is not exactly true that at half volume [HA]=[A-]. While exactly half of the acid was neutralized, concentration of A- is always a little bit higher than that of HA, as acid dissociates on its own. But for most weak acids in typical solutions (not too diluted) [HA]=[A-] is a quite good approximation.
4. (Original post by charco)
At the equivalence point all of the acid has reacted and produced salt.

Hence the moles of acid (originally) = moles of salt formed

When half the moles of acid (moles/2) have reacted they have also left behind half the moles of acid (moles/2) AND formed half the moles of salt (moles/2.

Hence moles of acid remaining = moles of salt formed

They are both in the same solution (volume must be the same), so:

[HA] = [A-]

At this point

[H+] = Ka

pH = pKa
Thank you
Could you explain q3 to me???

Posted from TSR Mobile
5. (Original post by Lamalam)
Thank you
Could you explain q3 to me???

Posted from TSR Mobile
While it is true that weak acids only dissociate by a small percentage, when the hydrogen ions on the RHS of the equilibrium react with a base they are removed from the equilibrium and the system must respond to make more by pushing the system to the RHS.

This process continues until all of the acid from the LHS has dissociated and the acid is completely used up.

Hence weak acids react stoichiometrically with bases.
6. (Original post by charco)
While it is true that weak acids only dissociate by a small percentage, when the hydrogen ions on the RHS of the equilibrium react with a base they are removed from the equilibrium and the system must respond to make more by pushing the system to the RHS.

This process continues until all of the acid from the LHS has dissociated and the acid is completely used up.

Hence weak acids react stoichiometrically with bases.
Thank you! )

Posted from TSR Mobile

TSR Support Team

We have a brilliant team of more than 60 Support Team members looking after discussions on The Student Room, helping to make it a fun, safe and useful place to hang out.

This forum is supported by:
Updated: November 28, 2013
Today on TSR

### The TSR A-level options discussion thread

Choosing A-levels is hard... we're here to help

### University open days

• Heriot-Watt University
School of Textiles and Design Undergraduate
Fri, 16 Nov '18
• University of Roehampton
Sat, 17 Nov '18
• Edge Hill University
Faculty of Health and Social Care Undergraduate
Sat, 17 Nov '18
Poll

The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE