Hedwigeeeee
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when we draw the titration curves.do we know exactly where the vertical points are or do we just estimate where they will be like if it is weak acid against strong base. then the ph in the veritical line should be around10 and final ph around 14?

I am really confused how to find the ph in equavalent points and what the final ph can reach.

like in this question. can we deduce the range of ph at equavalent point? or just estimate?Name:  IMG_20200926_215517-compressed.jpg.jpeg
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charco
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(Original post by Hedwigeeeee)
when we draw the titration curves.do we know exactly where the vertical points are or do we just estimate where they will be like if it is weak acid against strong base. then the ph in the veritical line should be around10 and final ph around 14?

I am really confused how to find the ph in equavalent points and what the final ph can reach.

like in this question. can we deduce the range of ph at equavalent point? or just estimate?Name:  IMG_20200926_215517-compressed.jpg.jpeg
Views: 16
Size:  109.8 KBName:  IMG_20200926_215526-compressed.jpg.jpeg
Views: 14
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pH is written just like that, pH, lower case p and upper case H.

The pH at the equivalence point can be obtained by dividing the inflection line into two equal sections. i.e. the mid-point of the vertical line.
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Hedwigeeeee
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(Original post by charco)
pH is written just like that, pH, lower case p and upper case H.

The pH at the equivalence point can be obtained by dividing the inflection line into two equal sections. i.e. the mid-point of the vertical line.
thanks. but how can we know what is the range of the vertical line? how to calculate it?
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charco
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(Original post by Hedwigeeeee)
thanks. but how can we know what is the range of the vertical line? how to calculate it?
If you want to calculate the pH at the equivalence point you don't need a graph. For a strong acid - strong base titration it is 7, the pH of the pure salt.

For weak acid - strong base titrations the equivalence point is the pH of the hydrolysed salt. This has ions that come from the base and the acid.
For 0.1 mol dm-3 sodium ethanoate (for example)
CH3COONa ==> CH3COO- + Na+
Ionic salts are completely dissociated into ions (sodium ions do not get hydrolysed, they are spectators to the following processes)

The ethanoate ion (the conjugate base of the weak acid) interacts with the water in the solution (hydrolysis) and sets up the following equilibrium:

CH3COO- + H2O <==> CH3COOH + OH-
This is the conjugate base hydrolysis equilibrium, and results in the formation of hydroxide ions. The hydrolysis constant, kh, is represented as follows:

kh = [CH3COOH][OH-]/[CH3COO-]

ka for ethanoic acid = 1.78 x 10-5 = [H+][CH3COO-]/[CH3COOH]
and
kw = [H+][OH-] = 1 x 10-14

If we divide kw by ka we get kh

Hence kh = 1 x 10-14/1.78 x 10-5 = 5.62 x 10-10

kh = [CH3COOH][OH-]/[CH3COO-] = 5.62 x 10-10

For 0.1 mol dm-3 sodium ethanoate solution, the ethanoate ions are 0.1 mol dm-3 (100% dissociation)

According to the stoichiometry of the hydrolysis equilibrium the [CH3COOH] = [OH-]

so we can write:

kh x 0.1 = [OH-]2

[OH-] = root(5.62 x 10-11)

[OH-] = 7.5 x 10-6

pOH = 5.13

pH = 8.87

/* ----------------------------------------------------------------------------*/

TL/DR

To find the pH of the hydrolysed salt divide kw by ka and solve for kh
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