suppress that Solve: message

Wrap it your Solve[...] expression with Quiet. So Quiet[Solve[...]] or Solve[...]//Quiet

why, after making a change to the code, I have to Evaluate the notebook twice for the changes to be reflected?

I'm just guessing, but it probably has to do with all these symbols being global and therefore the sequence of computation is important.

So, for example if at the top of my notebook a evaluate a=7 and then later evaluate b=a, at that point we have "facts" that a is 7 and b is 7. Now if you evaluate a=13, you've only changed one of those facts, i.e. a is now 13 but b is still 7. If instead you had written b := a, then b would "pick up" the value of a each time it is evaluated.

But I haven't gone through your notebook recently, so that's just a guess.

It would be nice if I could somehow suppress that Solve: message

Use exact numerical values

x = Solve[Rationalize[makeRelationEquation[Ka[pKa1], Kb[pKb], Ca1, Cb1, Va], 0], H];

Use Quiet with caution. It will suppress all messages.

With exact input the result is also exact. In this case Root objects. Instead of the complicated nested Part you can do this

sol = H /. x

Which will return a list of the 4 solutions in sol. To get the numerical (inexact values)

sol // N

To get exact representations of the Root objects

sol // ToRadicals

Paul,

Please post complete working code. relationEquation2 refers to symbols Ca and Cb that are not defined anywhere earlier in this thread. Here is what I evaluated

Kw = 1.00*10^-14;
Ka[pKa_] := 10^-pKa;
Kb[pKb_] := 10^-pKb;
pKa1 := 4.757;
pKb := 0.63;
Ca1 := 0.1;
Cb1 := 0.1;
Va := 10;
Vb := 10;
relationEquation = 
  Vb == Va*((Ca/(1 + H/Ka)) - H + Kw/H)/((Cb/(1 + Kw/(H*Kb))) + H - Kw/H);
makeRelationEquation[Ka_, Kb_, Ca_, Cb_, Va_] = relationEquation;
x = Solve[Rationalize[makeRelationEquation[Ka[pKa1], Kb[pKb], Ca1, Cb1, Va], 0], H];

Extract the solutions, select the positive one and convert to a numerical value

H /. x // Select[Positive] // First // N
(* 2.06037*10^-9 *)

Exact value

H /. x // Select[Positive] // First // ToRadicals
(* Long expression *)

/. is ReplaceAll.

I believe I may have a solution. From a chemistry perspective, the only valid solutions are those where H > 0. I believe that only one solution will meet the criterion H > 0 and although I could be wrong I propose to use that as a working hypothesis. In which case I can do this

x = Solve[makeRelationEquation[Ka[pKa1],Kb[pKb],Ca1,Cb1,Va],H,Assumptions->H>0];
x
pH = -Log10[Part[Part[Part[x,1],1],2]]

Do you have any other ideas? I have tested this with these values:

Kw = 1.00*10^-14;
Ka[pKa_] := 10^-pKa;
Kb[pKb_] := 10^-pKb;
pKb := 0.2;
Ca1 := 0.1;
Cb1 := 0.1;
Va := 10;
Vb := 10;
pKa := 0.63 and pKa := 4.757

Hi Eric, I am trying to adapt some of the code that you have previously provided to give me H (and hence pH) for just one value of Vb and this is what I have so far:

Kw = 1.00*10^-14;
Ka[pKa_] := 10^-pKa;
Kb[pKb_] := 10^-pKb;
pKa1 := 4.757;
pKb := 0.63;
Ca1 := 0.1;
Cb1 := 0.1;
Va := 10;
Vb := 10;
relationEquation=Vb==Va*((Ca/(1+H/Ka))-H+Kw/H)/((Cb/(1+Kw/(H*Kb)))+H-Kw/H);
makeRelationEquation[Ka_,Kb_,Ca_,Cb_,Va_]=relationEquation;
x = Solve[makeRelationEquation[Ka[pKa1],Kb[pKb],Ca1,Cb1,Va],H];
x
pH = -Log10[2.06037 * 10^-9] 

This gives the following:

Obviously I had to select which of the four values in x was the sensible one and I wondered if there is a better, more direct way of getting the result. Thanks.

That looks fantastic. At the moment I am working on getting data from a number of sources in order to compare and check results. This will take some time. Also, tomorrow I am taking delivery of a new computer and I'll need a few days to set it up and transfer stuff from my old computer.

All in all, I probably won't be able to look at this again until towards the end of next week. I hope that is OK with you and I really appreciate all the work you have put in so far.

I don't know if you are still keeping an eye on this discussion, Eric, but if you are I thought you might be interested in what I have finally come up with. If you, or anybody else, is interested I will happily attach my notebook (such as it is).

It's a first attempt and I am sure that you could pick it to pieces ..anyway you can find it here Trying to add a legend when using CombinePlots I couldn't have got even this far without your considerable help, so thank you once again.

If this is all true, - it is the truth, the whole truth and nothing but the truth

Yes, exactly, we can let the user manipulate Ka, Kb, Ca, Cb and Va

we want to show this in either orientation (i.e. for a target pH we want to see what Vb volume we need, or alternatively for a given Vb volume we want to know the pH we achieve

Yes again, but in both cases I would prefer pH on the y-axis

I had already sorted default values in Manipulate and that (and other things) are on the last notebook I attached and I did follow up on all the things you sent and referred to.

Okay, I think I understand that. What I'm confused about is what you're trying to find. Previously, it sounded like you were solving for pH, but now it sounds like you're solving for Ca. And when we started, we were solving for Vb. I guess I'm also confused about this: "we don't need to know anything about Ka or Kb". Either we have values for Ka & Kb, or we don't. They seem to be present in the formula you referenced, so I don't know how to set this up for the case where we don't know those values.

Here's what it sounds like to me. We have some acid solution. We must know some of its properties, but I'm confused about which ones. What we don't know but would like to know is its concentration, Ca. We have some base solutions, and we know some of their properties. If we can get to a neutral solution (I assume that's the "endpoint", but maybe it's some other pH) by adding a known amount of the base, we can calculate Ca with this complicated formula. So, we're solving for Ca, and we need a formula where Ca is defined in terms of all of these other independent variables. It sounds like the goal is to solve that complicated formula for Ca. Am I on the right track?

Sure. Sorry about the units (can't do superscripts here) Ca and Cb are the concentrations in mol dm-3 of an acid and a base Va and Vb are the volumes in cm3 of the acid and base Ka and Kb are the dissociation constants of the acid and base in mol dm-3 Kw is the ionic product of water in mol2 dm-6

The equation comes from an article in the journal School Science Review

Greatorex, D., 'Exact algorithms for acid/base titration curves', S.S.R. 1979, 214, 61, 94

I eventually got hold of a copy (attached) with help of a very kind ladyn at the Association for Science Education (the publishers of S.S.R.)

Attachments:

Greatorex, D., '...pdf

Vis-a-vis the divide by zero range error, it's a shame that instead of (for example) {Kb,0,1} which means 0 <= Kb <= 1 there could be way of stating the range as 0 < Kb <= 1

Thanks for that explanation - I had read up about SetDelayed but was not familiar with patterns. A couple of things arising:

1. I would like to use pKa instead of Ka, where pKa = -logKa (or Ka = 10^-Ka). I have attempted this by trying to follow the example used for pH but cannot get it to work. My attempt (commented out) is in the attached notebook.

2. I started my approach by calculating Vb for a range of values of pH because this seemed to be the quickest way to get started but what I really want to do is calculate pH for a range of values of Vb (this would more closely what happens in the laboratory). I would then plot pH on the y-axis against Vb on the x-axis. I am not sure whether we can use (some) of what we have already or would have to reformulate things to achieve this.

3. Manipulate is working nicely (as a newcomer to Mathematica, I am blown away by what can be done with it). I have attached Manipulate.PNG - in this screenshot, I don't really want to see the lower output because the upper input displays the result (11.3973) anyway. Is that possible? Also, can the number of decimal places in the result be specified so in this case, it would display as 11.40?

4. In manipulate, rather than specify a range that results in a slider control, is it possible (for some items at least) to offer the user a choice of three or four predefined values?

5. Lastly (for now, at least), is it possible to have the table output arranged as two columns rather than a collection of ordered-pair values?

I do hope I am not asking too much all at once. My revised notebook shows my attempts referred to above.

Attachments:

Manipulate.png

2.

...what I really want to do is calculate pH for a range of values of Vb...

If you can rewrite the equation in terms of ph = <some complicated expression>, then sure, that's doable. I'm not going to try to do it myself :). Then we'll just write a definition like ph[<arguments go here>]:=<new definition goes here>.

3B.

...Also can the number of decimal places in the result be specified...

Look at the various *Form methods: http://reference.wolfram.com/language/guide/DisplayOfNumbers.html

Try NumberForm, for example.

5.

is it possible to have the table output arranged as two columns

TableForm[Table[{x, Vb[x]}, {x, 3, 12.5, 0.2}]]

If you're going to want to do further processing on that data, then I suggest you first save the data in a variable:

dataPairs = Table[{x, Vb[x]}, {x, 3, 12.5, 0.2}];
TableForm[dataPairs]

The various *Form functions are just special wrappers that the front end knows how to interpret to give certain visualizations. You rarely want to use form-wrapped expressions in computations. In other words, you probably do NOT want to do this:

dataPairs = TableForm[Table[{x, Vb[x]}, {x, 3, 12.5, 0.2}]]

TableForm has a bunch of options if you want to tweak further.

The underscores are a pattern matching construct. Also, the := is SetDelayed, which is a common way to define a function.

http://reference.wolfram.com/language/ref/Blank.html http://reference.wolfram.com/language/ref/SetDelayed.html https://www.wolfram.com/language/elementary-introduction/2nd-ed/39-immediate-and-delayed-values.html https://www.wolfram.com/language/fast-introduction-for-programmers/en/function-definitions/