How can one calculate the pH of a solution?

So, I have 1 litre of acetate buffer, and 0.1 mol of oxonium ions/$\ce{H3O+}$ is added. The task is to calculate the pH of this solution.


What I was thinking:


$$\ce{CH3COOH + H2O <=> CH3COO- + H3O+}$$
$$\mathrm{pH} = \mathrm{p}K\_\mathrm{a} + \log\frac{c(\ce{Ac-})}{c(\ce{HAc})} = \mathrm{p}K\_\mathrm{a} + \log\frac{n(\ce{Ac-})}{n(\ce{HAc})}$$
$$\mathrm{pH} = 4.74 + \log\frac{n(\ce{Ac-}) - 0.1\ \mathrm{mol}}{n(\ce{HAc}) + 0.1\ \mathrm{mol}}$$


But neither $c(\ce{Ac-})$ nor $c(\ce{HAc})$ (which should, in theorey, be equal) is given. The only thing I know (or I think I know) is this: $$\frac{c(\ce{Ac-})}{c(\ce{HAc})} = \frac{1}{1} = 1$$


So, is it possible to calculate the pH from that? Am I missing something?




I'm tring to write the methodology  

If anything is wrong please point it out to me i'll be happy to correct!


To find $\small\ce{pH} $ of a solution you can: 


1. Find the conjugated base of your acid (for example $\small\ce{HCOOH}$ becomes $\small\ce{HCOO-}$ (correct me if i'm wrong here)
2. Write the equation of the solution  

Now you want to prepare an aqueous solution.
3. Find the concentration of $\small\ce{[CH\_3COOH]}$ 
you use the formula to determine the concentration in solution of acid


$$
 \ce{[CH\_3COOH]\ =\ C\ -\ [CH3COO- ]}
 $$


($\ce{C}$ is the concentration of brought acid in $\ce{mol/l}$)
4. Then you write that: 


$$\qquad\ce{[H+ ]\ =\ [CH3COO- ]}$$
as both dissociate from the same acid as 


$$\ce{CH3COOH\ ->\ CH3COO- \ +\ H+}$$
5. the $\small\ce{pH}$ concentration is given by:  

$$\ce{pH\ =\ -\log(\small[H\_3O+ ])}$$


$pH$ here you are! (without any unit)


 1

There are two things you need to know that are missing
1. The strength of the buffer (i.e. how many moles of Ac are in a litre of it. This includes undissociated and dissociated forms of accetate.
2. What is the pH of the buffer i.e. what does it measure before any NaOH is added. If the pH of the buffer is equal to the $pK\_a$ of the acid then indeed
$[\ce{HAc}] = [\ce{Ac^-}]$ ([·] symbolizes the molar concentration of ·). In order to solve the general problem we need to be able to calculate the fraction of the total Ac that is dissociated and the fraction that isn't.


The fraction that is dissociated comes right out of the Henderson - Hasselbalch equation and is $$f\_1 = 1/(1 + r\_1)$$ where $$r\_1 = 10^{(pH - pK\_1)}= [\ce{Ac^-}]/[\ce{HAc}]$$ as is clear from inspection of the Henderson - Hasselbalch equation. The subscript 1 indicates that $r\_1$ is the ratio of the number of acid ions that have lost 1 proton to the number that have lost $ 1 - 1 = 0$. In $f\_1$ the subscript is indicative that $f\_1$ is the fraction of the total Ac molecules that has become singly charged by loss of a single proton. When dealing with monoprotic acetic acid the subscripts aren't that important as there is only one proton to loose. But if the acid is polyprotic we have unionized, once ionized, twice ionized etc. ions to consider. The fractions of those ions are $f\_0$, $f\_1$, $f\_2$... and we have other ratios as well


$$r\_j = 10^{(pH - pK\_j)} = [H\_{n-j-1}Ac^{−j}]/[H\_{n-j}Ac^{-(j-1}]$$ Here "Ac" stands for the acid anion and n is the number of protons it can yield when fully dissociated. For acetic acid. $\ce{CH\_3COOH}$, and Ac is $\ce{CH\_3COO}$, n = 1 and j only has values of 0 or 1. With phosphoric acid, $\ce{H\_3(PO\_4)}$, Ac is $\ce{(PO\_4)}$, n = 3 and j = 0,1,2 or 3.


So suppose now we have a polyprotic acid with values for $r\_1$, $r\_2$, $r\_3$... and that there are x moles of the acid in a solution. Then there would be $xr\_1$ moles of the singly deprotonated species, $xr\_1r\_2$ moles of the doubly deprotonated, $xr\_1r\_2r\_3$ moles of the triply deprotonated and so on.Then the total number of moles of Ac would be the sum of the number of moles of each$$C\_{Ac} = x + xr\_1 +xr\_1r\_2 +xr\_1r\_2r\_3...$$ The fraction of the total that is undissociated is $$f\_0 = x/(x + xr\_1 +xr\_1r\_2 +xr\_1r\_2r\_3...) = 1/(1 + r\_1 +r\_1r\_2 + r\_1r\_2r\_3...)$$ The fraction that is singly dissociated is $r\_1$ times this $$f\_1 = r\_1f\_0$$ and the fraction that is doubly dissociated is $r\_2$ times that $$f\_2 = r\_2f\_1$$ and, in general $$f\_j = r\_jf\_{(j-1)}$$ 


At this point let's remember that $f\_j$ is a function of the solution pH and all the pK's of the acid in question and that it is the fraction of the anions of that acid that carry charge -j. Thus we can write an expression for the total charge on all species of Ac at a given pH. This is $$Q\_{Ac} = -C\_{Ac}(0f\_0 + 1f\_1 + 2f\_2 + ...)$$


Before going on to show you how Q solves buffering problems let me stop to suggest that the simplest way to work with it is to make an Excel (or other) spread sheet. Designate a column for pKs and a cell into which the pH goes. For example, put pH into cell A1 and start the pKs list in A2, Then in B2 put =10^($A$1 - A2). Using $A$1 lets you copy and paste B2 into as many cells as you have pKs. The B column now contains the r corresponding to the pK in the cell to the left of it. Now enter the formula for $f\_0$ in a cell and make another column with the $f\_j = r\_jf\_{(j-1)}$ formula in it. So how many pK's. I say make the spread sheet for 5 or 6. Why? Well lets go back to $\ce{CH\_3COOH}$ for a minute. It doesn't have 1 proton to give, as we have been assuming. It actually has 4. Are the other three ever coming off? Not with any base in my lab but we can model those other protons simply by assigning pKs that are so high (say 50) that the f values for anything other than $f\_0$ or $f\_1$ are 0. The point being that if you are going to go to the trouble to make the spreadsheet you might as well make it big enough to handle any acid you may ever encounter as it easily handles anything up to its maximum size using this trick.


Now how to use $Q(pH)$. If there are a total of $C\_{Ac}$ moles of Ac in a solution the negative charge on them at $pH\_0$ is $C\_{Ac}Q(pH\_0)$. At $pH\_1$ it is $C\_{Ac}Q(pH\_1)$. Thus to move the solution from $pH\_0$ to $pH\_1$ you must supply or remove charge of $$\Delta Q\_{Ac}(pH\_0\ce{->}pH\_1) = C\_{Ac}Q(pH\_1) - C\_{Ac}Q(pH\_0)$$ by adding or absorbing protons. If $ pH\_0 > pH\_1$ then $Q(pH\_1) > Q(pH\_0)$ (less negative) and so the difference will be positive indicating that protons (acid) will need to be added to effect this pH shift.


The acid species in the solution are not the only thing that emits or absorbs protons when pH changes. The solvent does too. 


$$\Delta Q\_{W}(pH\_0\ce{->}pH\_1) = 10^{-pH\_1} - 10^{-pH\_0} + (10^{(pH\_0 - pK\_w)} - 10^{(pH\_1 - pK\_w)})$$ represents the number of protons that must be supplied (or absorbed) to change the pH of water from $pH\_0$ to $pH\_1$.


Now let's use this to solve the original question. We'll assume that someone added 0.2 M acetic acid to a flask containing some water, added NaOH solution until the pH was 4.76, added water to near 1 L, trimmed with NaOH and HCl and then made up to the 1L mark. At this point we have a liter of 4.76 buffer with $C\_{Ac}$ = 0.2 mol ( 0.1 mol of Ac and 0.1 mol of $Ac^{-1}$. This implies that $f\_0 = f\_1 = 0.5$ Also
$$\Delta Q\_{Ac}(pH\_0\ce{->}pH\_1) = 0.02(Q(pH\_1) - Q(4.76))$$
$$\Delta Q\_{W}(pH\_0\ce{->}pH\_1) = 10^{-pH\_1} - 10^{-4.76} + (10^{(4.76 - pK\_w)} - 10^{(pH\_1 - pK\_w)})$$


We know that if by adding 0.1 M protons to the liter we cause it to settle at $pH\_1$ then $$\Delta Q\_{Ac}(4.76\ce{->}pH\_1) + \Delta Q\_{W}(4.76\ce{->}pH\_1) = 0.1$$ which simply says that the protons we added have to be equal to the protons absorbed. Or, put another way, the net protons transferred has to be 0


$$\Delta Q\_{Ac}(4.76\ce{->}pH\_1) + \Delta Q\_{W}(4.76\ce{->}pH\_1) - 0.1 = 0$$ The job now becomes one of finding $pH\_1$ which satisfies this equation. That is actually quite easy to do. If you enter the formulae given here into a spreadsheet and set it up to compute the left hand side of this last equation you can then try various values of $pH\_1$ until you find one that gives a result close to 0. Or you can let Excel's solver do it for you. It didn't take me long to figure out that adding 0.1 mol protons to 1 L of an 0.2 M acetate buffer at 4.76 would lower its pH to 2.74.


Now this post has gone on long enough so I won't have space to tell you about extension of this technique to much more complicated systems. You can use it to find the pH of complicated mixes of weak and strong acids and bases or any materials which have buffering capacity. I ginned it up to predict the pH of brewer's mash which is a mixture of weak acid (bicarbonate) and several malts (each of which has its own buffering properties).


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