[cliverlong]

I have realised I don't really understand the basics of acid-base calculations and I need some clarification.

I will consider the titration of 100cm³ of 0.1M NaOH against 100cm³ of  0.1M HCl (or is that HCl against NaOH ??) . I am adding alkali to acid

Question 1

Now as HCl is a strong acid I assume it is fully dissociated in aqueous solution

HCl + H₂O → Cl^- + H₃O⁺

so pH of the acid is calculated from

-log[H₃O⁺]

Since HCl is fully dissociated I say [H₃O⁺] = 0.1 mol dm^-3

so -log[0.1] = 1 (pH of the strong acid)

If I try to calculate the Ka

Ka = [H₃O⁺][Cl^-] / [HCl]

but what value do I put in for [HCl] if it is fully dissociated? Zero? Then I have an infinite Ka for HCl.

Question 2

Now I consider adding the 100cm³ (0.1dm³) of the alkali to the 100cm³ (0.1dm³) of the acid.

Since the strong alkali is fully dissociated and the quantities (moles) of H⁺ and OH^- are the same they neutralise, or combine to form water.

How do I calculate the pH of the resulting solution if the H⁺ has been “used up” by the OH^- ? I feel Kw = Ka x Kb is relevant here but I can't think how to apply it.

I'm thinking the dissociation of water

2H₂O  H₃O⁺ + OH^-

becomes an important source of H₃O⁺ + OH^- ions at the equivalence point but I can't think how to use this to calculate the pH (which should be 7).

Ta

Clive

[Schrödinger]

Question 1:

When you say that HCl is completely dissociated, you write an irreversible reaction as follows.

HCl + H₂O → Cl^- + H₃O⁺

But it is important to keep in mind that K_a is defined as the equilibrium constant for the reversible reaction of acid ionization.

According to your argument, you are actually trying to calculate how reversible a completely irreversible reaction is. So, K_a is infinity.

This is quite clear, I presume?

[Borek]

2H₂O  H₃O⁺ + OH^-

Kw=[H⁺][OH^-]

[cliverlong]

Question 1:

When you say that HCl is completely dissociated, you write an irreversible reaction as follows.

HCl + H₂O → Cl^- + H₃O⁺

But it is important to keep in mind that K_a is defined as the equilibrium constant for the reversible reaction of acid ionization.

According to your argument, you are actually trying to calculate how reversible a completely irreversible reaction is. So, K_a is infinity.

This is quite clear, I presume?

Indeed Watson.

I'll have another go assuming HCl is 99% dissociated (enough for a strong acid I'm guessing)

I'm trying to see if I can plot the titration curve by calculating pH (using spreadsheet) as I add increments of strong alkali

I'm guessing a problem occurs near equivalence point due to contribution of dissociation of water to level of H⁺. I haven't in my mind worked out how to handle the conflicting processes of OH^- absorbing the H⁺ to make H₂O and simultaneously the H₂O splitting up again - but I'm sure I will make some progress. I'll post again when I get stuck.

Clive

[Borek]

pKa for HCl is estimated as about -4

Note that if there is no other sources of H⁺ & OH^- [H⁺] = [OH^-].

You may also visit

http://www.chembuddy.com/?left=pH-calculation&right=pH-strong-acid-base

[cliverlong]

pKa for HCl is estimated as about -4

I found a dissociation of 99% gave a Ka of around 1.
A dissociation of 99.9% gave a Ka of around 10

So pKa of -4 suggest dissociation around 99.99999%

By similar calculation I get Kb of around 10 if I assume NaOH is 99.9% dissociated.

Note that if there is no other sources of H⁺ & OH^- [H⁺] = [OH^-].

You may also visit

http://www.chembuddy.com/?left=pH-calculation&right=pH-strong-acid-base

I don't follow what the term C_a is .

When I did a few calculation using

100cm³ (0.1 dm³) of 99.9% dissociated HCl and calculated the resulting pH when I added amounts of (99.9% dissociated) NaOH I took the following approach

Total moles H+ from HCl (most of the time H⁺ from H₂O too small to affect calculation) - Total moles added OH^- from NaOH (again almost all calculations OH^- from H₂O too small to affect calc)

Then calculated resulting pH

99cm³ NaOH  pH = 3

Even when I calculated using 99.9cm³ of NaOH added and took into account the moles of H⁺ from water I still get pH = 4.3 to 2 sig. figs. of resulting solution. This just seems too small to me. Surely disociation H₂O should being having an impact at these volumes?

What is wrong in my approach?

Thanks

Clive

[Borek]

By similar calculation I get Kb of around 10 if I assume NaOH is 99.9% dissociated.

No need to guess:

http://www.chembuddy.com/?left=BATE&right=dissociation_constants
http://www.chembuddy.com/?left=FAQ

I don't follow what the term C_a is .

http://www.chembuddy.com/?left=pH-calculation&right=symbols

What is wrong in my approach?

Nothing yet. Compare tables here:

http://www.titrations.info/acid-base-titration-end-point-detection

[cliverlong]

OK I get it now.

My theoretical titration plots (using spreadsheet) are much closer to those for the various strong/weak combinations given in textbooks and online.

Basically for strong acids and alkalis, the contribution of water ionization to H+ and OH- is insignificant until quantities of H+ and OH- from the acid and alkali are almost exactly equal. This leads to the steep part of the titration curve near the equivalence point.

Conversely for weak acids and alkalis, water ionization provides a significant source of H+ and OH- although the range of pH change during titration is much narrower. Hence a "flatter" titration curve.

I hadn't appreciated how big is the range in Ka / Kb between strong and weak acids/alkalis.

Clive

[Borek]

My theoretical titration plots (using spreadsheet) are much closer to those for the various strong/weak combinations given in textbooks and online.

Try my BATE (pH calculator) - it has a free 30 days trial and you can use without any obligations. It shows titration curves on the fly, so you can play with various combinations of Ka/Kb values.