# Finding pH of water

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Hi All,

This should be easy, but I can't seem to find it on the 'net.

I'm trying to get my head around calculating the pH of a solution given its chemical makeup. I had hoped that starting with something easy, like water, would help me.

But all the research and attempted calculations don't add up.

The equilibrium equation of water is Kw mol2 dm-6 = [H3O+]mol dm-3 [OH-]mol dm-3

I'm told that Kw is 1.0 x 10-14. (Why? Where did this come from? Was it just made up because the pH scale only goes to 14??)

I've calculated [H3O+] to be 19.02322 g/mol or 0.01902332 mol dm-3 and [OH-] is 17.00734 g/mol or 0.01700734 mol dm-3

First, do I have the correct values for [H3O+] and [OH-]?

Secondly, I can't balance the the equation as 10-14 ≠ 0.0003235360711688!

Where am I going wrong?

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10 minutes ago, NotYou said:

Hi All,

This should be easy, but I can't seem to find it on the 'net.

I'm trying to get my head around calculating the pH of a solution given its chemical makeup. I had hoped that starting with something easy, like water, would help me.

But all the research and attempted calculations don't add up.

The equilibrium equation of water is Kw mol2 dm-6 = [H3O+]mol dm-3 [OH-]mol dm-3

I'm told that Kw is 1.0 x 10-14. (Why? Where did this come from? Was it just made up because the pH scale only goes to 14??)

I've calculated [H3O+] to be 19.02322 g/mol or 0.01902332 mol dm-3 and [OH-] is 17.00734 g/mol or 0.01700734 mol dm-3

First, do I have the correct values for [H3O+] and [OH-]?

Secondly, I can't balance the the equation as 10-14 ≠ 0.0003235360711688!

Where am I going wrong?

From measurement, [H3O+] and [OH-] are each 10⁻⁷ mol/l in pure water at room temp, hence you get a pH of 7 and an ionic product, Kw of 10⁻¹⁴.

It looks as if you have calculated the number of moles of pure H3O+and OH- you would have in a litre, if water were 100% ionised.

But, as it is only very slightly ionised, so you will get wrong numbers that way.

At least, this is how it looks to me at first glance.

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Posted (edited)
9 hours ago, NotYou said:

This should be easy, but I can't seem to find it on the 'net.

I think you said you are studying on your own, not following any formal course so I will lay it out in some detail as you do seem to have got in a pickle.

9 hours ago, NotYou said:

The equilibrium equation of water is Kw mol2 dm-6 = [H3O+]mol dm-3 [OH-]mol dm-3

No that is not the equilibrium equation so let us start from there.

For any chemical reaction we start with reactants say A and B and end up with products say C and D.

We write this as

$A + B \leftrightarrow C + D$

Note there are no square brackets in the chemical reaction, but there is a double headed arrow because every chemical reaction is really two reactions one going forward from reactants to products and one going in the reverse direction.

Now every chemical reaction also has a speed or velocity.

In general this speed is proportional to the concentrations of the reactants.

So for the forward reaction (note the single arrow this time) we have

$A + B \to C + D$

With velocity Vforward given by

${V_{forward}} = {k_{forward}}\left[ A \right]\left[ B \right]$

Where kforward is the combined constant of proportionality for both reactants.
Note also we have now introduced the square brackets to represent the reactant concentrations which we said the speeds were proportional to.

Similarly we can create the corresponding equations for the reverse reaction (note the reversed arrow)

$A + B \leftarrow C + D$

and

${V_{reverse}} = {k_{reverse}}\left[ C \right]\left[ D \right]$

Now we ask what chemical equilibrium means.
It means that the speed of the forward reaction equals the speed of the reverse reaction so there is no overall change in concentrations of reactants or products.

${V_{forward}} = {V_{reverse}}$

Which may be written

$\frac{{{V_{forward}}}}{{{V_{reverse}}}} = 1$

Substituting in the expressions for these speeds,

$\frac{{{k_{forward}}\left[ A \right]\left[ B \right]}}{{{k_{reverse}}\left[ C \right]\left[ D \right]}} = 1$

Rearranging to form the ratio of proportionality constants

$\frac{{{k_{forward}}}}{{{k_{reverse}}}} = \frac{{\left[ C \right]\left[ D \right]}}{{\left[ A \right]\left[ B \right]}}$

We arrive at the equilibrium constant for this particular reaction

${K_{equilibrium}} = \frac{{{k_{forward}}}}{{{k_{reverse}}}} = \frac{{\left[ C \right]\left[ D \right]}}{{\left[ A \right]\left[ B \right]}}$

Note that because of the rearrangement the products are on top of the fraction and the reactants on the bottom!

This leads to the mnemonic rule

$\frac{{{\rm{Products}}}}{{{\rm{Reactants}}}}$

When using these constants.

OK so to use all that for the reaction which is the self ionisation of water.

First write the reaction

${H_2}O + {H_2}O \leftrightarrow {H_3}{O^ + } + O{H^ - }$

At this point I am going to cheat and simplify because using the hydroxionium ion may be strictly more correct, but it complicates the maths so I will just use the single hydrogen ion and single hydroxyl ion, as I want to proceed to pH issues.

${H_2}O \leftrightarrow {H^ + } + O{H^ - }$

So we have lost B from the reactants.

So the equilibrium constant for this equation is

${K_{equilibrium}} = \frac{{\left[ {{H^ + }} \right]\left[ {O{H^ - }} \right]}}{{\left[ {{H_2}O} \right]}} = 1.8x{10^{ - 16}}$

Note these are always experimentally determined values. They cannot be found from theory alone.

Now the concentration of the water in dilute aqueous solutions is constant and given by

$\left[ {{H_2}O} \right] = \frac{{1000g/litre}}{{18g/mole}} = 55.5moles/litre$

This gives us a new equation

$\frac{{\left[ {{H^ + }} \right]\left[ {O{H^ - }} \right]}}{{55.5}} = 1.8x{10^{ - 16}}$

or

$\left[ {{H^ + }} \right]\left[ {O{H^ - }} \right] = 1.8x{10^{ - 16}}x55.5 = 1.0x{10^{ - 14}}$

Which is the value you have come across and is called the ionic product constant for water

This is given the symbol Kw to distinguish it from the equilibrium constant.

$\left[ {{H^ + }} \right]\left[ {O{H^ - }} \right] = 1.8x{10^{ - 16}}x55.5 = {K_w} = 1.0x{10^{ - 14}}$

Noting further that the concentration of hydrogen ions must equal the concentration of hydroxyl ions we have

$\left[ {{H^ + }} \right] = \left[ {O{H^ - }} \right] = \sqrt {{K_w}} = 1.0x{10^{ - 7}}$

So to pH

It is sometimes convenient to take the logarithm (to the base 10, they are not natural logs) and for any X ( which could be a suitable concentration or constant)

the p function of X (spoken pX all one word)  is defined as

pX = -logX

So for our water

pH = -log[H+]

pOH = -log[OH-]

So

-log[H+] -log[OH-]  =  log (10-14) = -14

pH + pOH = 14 =pKw

Therefore in a neutral solution or pure water

pH =  pOH  = 7

You need to fully understand all this before trying to calculate the pH after introducing another substance.

How are we doing ?

Edited by studiot
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12 hours ago, studiot said:

Note these are always experimentally determined values. They cannot be found from theory alone.

So it's impossible to find the pH of a solution given only the molecular concentrations?

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20 minutes ago, NotYou said:

So it's impossible to find the pH of a solution given only the molecular concentrations?

That wasn't quite what I (or exchemist) said.

There are many constants in Science that can only be determined by experiment as we have no deeper theory to derive them from.

These include the mass of the electron, the acceleration due to gravity, Avogadro's number, the thermodynamic constant gamma to name a few and show how widespread they are.

However once they have been measured, they can be used in conjunction with theory to determine other desired quantities.

The ionisation constant for water is just one of these.

Soulubilities (solubility constants) are very similar.

That is also why engineers and scientists have extensive tables of Physical and Chemical Constants and other data.

I posted a lot of material

Look carefully again at the part where I started to to develop the pH scale and ask questions about what you don't follow.

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Posted (edited)
1 hour ago, NotYou said:

So it's impossible to find the pH of a solution given only the molecular concentrations?

If you know the concentration of H+ or H3O+, you are home and dry. You take the log of the value and take the negative of what you get (the logs are almost invariably -ve, so this procedure gives you a +ve value for the pH.)

If you want to find the pH without knowing the concentration of H+ (or H3O+), you need to know the dissociation constant for the species present. For pure water we know this, of course. For other solutions, it depends what you've got. They are documented for many molecules. For example you can find values of Ka for all the commonly encountered acids.

It is also possible to calculate equilibrium constants, in principle including dissociation constants, from the thermodynamics of the species involved: ΔG = -RT lnK, where ΔG is the change in Gibbs free energy for the dissociation.

Edited by exchemist
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This is where I get confused.

Given:

[H+][OH−]/[H2O]=1.8x10−16
[H+]=[OH−]=1.0x10−7

And:

[H] = 992.1 m/l
[HO] = 58.8 m/l
[H2O] = 55.5 m/l

How can:

(992.1*58.8)/55.5 = 1.8x10−16 ?
992.1 = 58.8 = 1.0x10−7 ?

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Posted (edited)
7 hours ago, NotYou said:

This is where I get confused.

Given:

[H+][OH−]/[H2O]=1.8x10−16
[H+]=[OH−]=1.0x10−7

And:

[H] = 992.1 m/l
[HO] = 58.8 m/l
[H2O] = 55.5 m/l

How can:

(992.1*58.8)/55.5 = 1.8x10−16 ?
992.1 = 58.8 = 1.0x10−7 ?

OK, you still have the same confusion as originally, it seems.

In my original reply to you, I mentioned that you seemed to be treating water as fully ionised, when it is in fact barely ionised at all. It looks as if we need to go back and revisit what dissociation into ions involves. (Forgive me if you know all this, but it looks from your posts as if you may not.)

Water consists of H-O-H molecules, right? H is bonded to O, by means of a covalent bond. You do not have lots of free "H" and "OH" floating around.

However, what happens to a tiny fraction of the molecules is that they split, or dissociate, into ions. The covalent bond involves two electrons, one from H and one from O, being shared between the two atoms. In the dissociation of a water molecule, both electrons go to the O atom, leaving H without an electron. So you get H+, because it is one electron short and OH-, because it has one extra.

In bulk water, there is a dynamic equilibrium, in which molecules are continually dissociating into ions and ions are continually recombining into neutral molecules again. But, and this is important to understand, the dissociated state requires more energy than the neutral state, so only a tiny fraction of the molecules are dissociated into ions at any given moment.

Whereas what you have been doing, in effect, is counting all the H atoms present, bonded or not, towards your "[H]" figure. That's why it's wrong.

You need to know the proportion of the molecules that is ionised at any given moment. This is extra information. You can't calculate it just from the chemical formula You get this from Kw, which is an experimentally determined figure.

Edited by exchemist
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2 hours ago, exchemist said:

You can't calculate it just from the chemical formula You get this from Kw, which is an experimentally determined figure.

Again: So it's impossible to find the pH of a solution given only the molecular concentrations!

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3 minutes ago, NotYou said:

Again: So it's impossible to find the pH of a solution given only the molecular concentrations!

Again NO.

You have been told at least four times now that this is not true.

However until you understand how to calculate the pH of pure water with nothing in solution, you will never be able to understand the next step of how to calculate the pH of a solution.

If you have a sample of pure water, it does not matter whether this is 1mL in a pipette (or even 0.1mL) or 1 million gallons in a tank the very small proportion that is ionised to H+ and OH- is the same.

Do you understand this  ?

2 hours ago, exchemist said:

Water consists of H-O-H molecules, right? H is bonded to O, by means of a covalent bond. You do not have lots of free "H" and "OH" floating around.

However, what happens to a tiny fraction of the molecules is that they split, or dissociate, into ions.

I thought that was pretty clear.

They say it takes a good teacher to spot what a student has done wrong and put that right, rather than just say "you have done this the wrong way  -  do it this different way"

+1

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Posted (edited)
3 hours ago, NotYou said:

Again: So it's impossible to find the pH of a solution given only the molecular concentrations!

No, that's wrong. You can, if you know the dissociation constant as well. (A dissociation constant is just the equilibrium constant for a dissociation reaction.)

You can't know the pH just from the chemical equation for the dissociation, since that does not tell you where the equilibrium lies between the left hand side and the right hand side. That's the missing piece of information that the dissociation constant (or the ionic product in the case of water) tells you.

To make life easier for you, if you have a strong acid, such as HCl or HNO3, you can assume that it is fully dissociated (unless you are dealing with very high concentrations). So for these, if you know the concentration of acid from the amount you added, you can set [H+] equal to that, since every molecule of acid gives you one H+ ion in solution.  So knowing the acid concentration you can just work out the pH from that.

If you have a weak acid, like acetic acid, then you need to look up Ka for the acid and work out [H+] using that, knowing the concentration of acid you have added.

If you have a more complex mixture, involving and acid and a base together, then if it is a strong acid with a strong base, you can assume full neutralisation occurs and that the pH will be determined just by what is left over. If you have a weak acid, or a weak base, then you will need to know the equilibrium constant for the neutralisation reaction involved.

So it is doable in all these cases, so long as you know the molecular concentrations AND the relevant equilibrium constant.

Edited by exchemist
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On 6/2/2021 at 11:47 PM, exchemist said:

You can, if you know the dissociation constant as well.

Are these dissociation constants available for everything? Where would I find a list of them?? (Please don't say "Google it")

On 6/2/2021 at 8:13 PM, studiot said:

very small proportion that is ionised

So that would mean that these dissociation constants are dependant on temperature and pressure? Is there a formula for that/them??

On 6/2/2021 at 11:47 PM, exchemist said:

if you have a strong acid, such as HCl or HNO3, you can assume that it is fully dissociated (unless you are dealing with very high concentrations)

What happens when either are pure? Can they be pure??

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6 minutes ago, NotYou said:

Where would I find a list of them?? (Please don't say "Google it")

Why not?

6 minutes ago, NotYou said:

So that would mean that these dissociation constants are dependant on temperature and pressure?

The temperature dependence is usually quite strong; the pressure has less effect.

7 minutes ago, NotYou said:

Is there a formula for that/them??

Yes, but it won't help much because you need to know the energy change for the reaction.

9 minutes ago, NotYou said:

What happens when either are pure? Can they be pure??

Pure HCl is a gas.
The case of pure liquid acids like H2SO4 or HNO3 is complicated.

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1 hour ago, NotYou said:

Are these dissociation constants available for everything? Where would I find a list of them?? (Please don't say "Google it")

So that would mean that these dissociation constants are dependant on temperature and pressure? Is there a formula for that/them??

What happens when either are pure? Can they be pure??

Yes the dissociation constants are known for thousands of acidic compounds. OK I won't say google them, but the fact is you need to look them up somewhere.

When I was doing this sort of thing back in the day, we used what was known as the "Rubber Book", a huge telephone directory size book that sat on the bookshelf in every chemical laboratory. Details here: https://en.wikipedia.org/wiki/CRC_Handbook_of_Chemistry_and_Physics

I have no doubt that similar compendia of this data exist in on-line form. Checking briefly on-line just now, it looks as though most of the tables one finds actually still reference the Rubber Book.

Re temperature dependence, I gave a formula for how equilibrium constants depend on temperature, in an earlier post: ΔG = -RTlnK. The snag, as @John Cuthberpoints out, is you need to know ΔG for the reaction in question, which means looking up the relevant thermodynamic data for the reaction. You will find this ......in the Rubber Book.

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3 hours ago, exchemist said:

You will find this ......in the Rubber Book.

For those who aren't aware of it...
https://en.wikipedia.org/wiki/CRC_Handbook_of_Chemistry_and_Physics

But, not every possible reaction is in there.
So, fundamentally, if you produce (or find) a new compound, you might be able to guess it's properties but generally, you will have to measure them.

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1 minute ago, John Cuthber said:

For those who aren't aware of it...
https://en.wikipedia.org/wiki/CRC_Handbook_of_Chemistry_and_Physics

But, not every possible reaction is in there.
So, fundamentally, if you produce (or find) a new compound, you might be able to guess it's properties but generally, you will have to measure them.

Er yes I put that link in my post....😃

But tell me, does CRC or anyone else maintain an on-line version of the Rubber Book? It seems like an obvious thing to put into a web-accessible database.

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Posted (edited)
9 hours ago, NotYou said:

So that would mean that these dissociation constants are dependant on temperature and pressure? Is there a formula for that/them??

Since this is all you seem to want to know, the dependence of pressure is negligable for range of pressures encountered outside specialist facilities.

Here is a table of dependence on temperature.

As can been seen this varies by a factor of about 100 over the range 0oC to 60oC

Both temperature and pressure dependence are experimental facts, which is why everyone is talking about tables of data.

9 hours ago, NotYou said:

Are these dissociation constants available for everything? Where would I find a list of them??

I would suggest looking in the Chemistry or Science reference section of your local library.

Since you are in Australia, they will probably have the latest copy of

Kaye and Laby, Tables of Physical and Chemical Constants from the UK National Physical Laboratory

or

Lange's handbook of Chemistry

or

The CRC (rubber) book

Rather smaller books of  tables prepared for national schools and colleges, such as the Nuffield one or SI Chemical Data by Aylward and Findlay (Wiley)

But you need to go right back to my proforma standard equation A + B = C + D to know what to look for.

You would also need this equation to use the information correctly, as exchemist has already indicated.

There are four different possibilities

Strong acid + strong base

Strong acid plus weak base

Weak acid plus strong base

Weak acid plus weak base.

Each of these leads to a different equation, which must be handled appropriately (differently).

The last one is often used in a simplified form called the Henderson - Hasselbalch equation, (which itself has two versions) especially by the life sciences (biochemistry etc) where is was originally stated.

So you see there is not one equation but getting on for 10 different ones depending upon circumstances and complexities and where my express version was heading.

I am sorry you didn't like it as it involved me in substantial work on your behalf.

Edited by studiot
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On 6/5/2021 at 2:51 AM, exchemist said:

But tell me, does CRC or anyone else maintain an on-line version of the Rubber Book? It seems like an obvious thing to put into a web-accessible database.

I found it as PDF's at analysischamp.com/links2.html

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Posted (edited)
3 hours ago, NotYou said:

I found it as PDF's at analysischamp.com/links2.html

Hmm, well done...but this looks like an amateur site and possibly pirated. I wonder if the site owner has got copyright licence from CRC to do this.

Edited by exchemist
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Posted (edited)
On 6/1/2021 at 7:35 AM, studiot said:

Vforward = Vreverse

Which may be written

Vforward
Vreverse = 1

It can also be written as:

1 = Vreverse
Vforward

Which becomes:

Kequilibrium = [A][ B] = kreverse
[C][D]     kforward

How will that change the rest of the proof?

Edited by NotYou
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On 6/1/2021 at 7:35 AM, studiot said:

Vforward = Vreverse

What happens if VforwardVrevers??

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25 minutes ago, NotYou said:

It can also be written as:

1 = Vreverse
Vforward

Which becomes:

Kequilibrium = [A][ B] = kreverse
[C][D]     kforward

How will that change the rest of the proof?

It doesn't change anything, it is not a proof.

It is a derivation to show where it comes from.

It also shows that you are correct a little bit of information is lost in the derivation process and it is a matter of convention which way up we define the fraction.

This is an arbitrary choice that is internationally adopted and must be simple remembered.

I think I made the comment that remembering which way up trips many students up so stressed this point.

The convention adopted does has the advantage that these constants are very small (much less than 1).

So this makes the pH and pX equations fit neatly into a convenient range of numbers.

When you start to introduce other substances to the pure water this changes the pH and other constants become involved.

I have starred the comment in the attachment below.

So it becomes even more important to get the  fractions the right way up.

Here is a very simple calculation for the pH of 0.03M HCl in pure water.

Here you also need to know that HCl is a 'strong acid'

This means that it is totally dissociated in water.

So the concentration from the hydrogen chloride of H+ ions = concentraction of Cl- ions = 3 x 10-2M

We know that the concentration of H+ ions from the water is 1 x 10-7M

So the total concentration of H+ ions is (0.03 + 0.0000001)M = .0300001M

So we ignore the 0.0000001M from the water.

So the log10 of 0.03 = log (3) + log (10-2) = 0.48 + (-2.0) = -1.52

So the pH of 0.03 HCL is -(-1.52) = +1.52

This is the simplest calculation and shows what happens when either a source/sink of H+ or OH- ions is added.

In this case the OH- concentraction is unaffected since Cl- ions are added.

Note that this solution is once again electrically neutral (contains the same number of + and - charges)

and at chemical equilibrium.

So if you added 36.5 x.03 = 1.1 grammes of HCl to every litre of pure water the pH would be 1.52

So as the simplest possible exercise can you predict what would happen to the pH of pure water if you added 5.84 grammes of sodium chloride (NaCl or common salt) to pure water  ?

6 minutes ago, NotYou said:

What happens if VforwardVrevers??

Then the solution would not be in chemical equilibrium and this means that the concentrations would be changing over time.

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2 minutes ago, studiot said:

it is not a proof.

It is a derivation to show where it comes from.

Sorry, didn't know what else to call it.

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1 minute ago, NotYou said:

Sorry, didn't know what else to call it.

Science in general does not do 'proofs'  - That is for Mathematicians and Lawyers, although their definitions of the word are somewhat different.

Science does hypotheses and deductions, which can tested against observations.

But it should always be open to modification following further observations which show something different.

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9 minutes ago, studiot said:

Here is a very simple calculation for the pH of 0.03M HCl in pure water.

Thanks for this extra info, Studiot. I haven't got there yet, but that is what I am looking to learn (and more).

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